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All questions of Permutations and Counting for Grade 12 Exam
The number of ways in which the 6 faces of a cube can be painted with 6 different colours is- a)6
- b)12
- c)6!
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
The number of ways in which the 6 faces of a cube can be painted with 6 different colours is
a)
6
b)
12
c)
6!
d)
none of these
|
Preeti Iyer answered |
Consider the face that has red color. There are five face
In how many ways can a cricket team of 11 players be chosen out from a squad of 14 players, if 5 particular players are always chosen?- a)84
- b)144
- c)1001
- d)1365
Correct answer is option 'A'. Can you explain this answer?
In how many ways can a cricket team of 11 players be chosen out from a squad of 14 players, if 5 particular players are always chosen?
a)
84
b)
144
c)
1001
d)
1365
|
Aryan Khanna answered |
Total no of players = 14 out of which 5 are fixed.
So, 11-5 = 6
Remaining players = 14 - 6
= 9 players
9C6 = 9!/(3!*6!)
= 84
So, 11-5 = 6
Remaining players = 14 - 6
= 9 players
9C6 = 9!/(3!*6!)
= 84
Four alphabets A, M, P, O are purchased from a warehouse. How many ordered pairs of initials can be formed using these?- a)10
- b)16
- c)18
- d)12
Correct answer is option 'D'. Can you explain this answer?
Four alphabets A, M, P, O are purchased from a warehouse. How many ordered pairs of initials can be formed using these?
a)
10
b)
16
c)
18
d)
12
|
Riya Banerjee answered |
Total number of letters = 4
Number of ordered pairs of letters that can be formed like (A, M) or (P, O) etc = 4P2 = 4!/2!
= 24/2
= 12
Number of ordered pairs of letters that can be formed like (A, M) or (P, O) etc = 4P2 = 4!/2!
= 24/2
= 12
?
In how many ways can 3 letters be posted in 4 letter boxes?- a)27
- b)4!
- c)64
- d)3!
Correct answer is option 'C'. Can you explain this answer?
In how many ways can 3 letters be posted in 4 letter boxes?
a)
27
b)
4!
c)
64
d)
3!
|
|
Preeti Iyer answered |
The first letter can be posted in 4 ways. So, total outcomes about the first letter = 4.
For every outcome about the first letter, the second letter can be posted in 4 ways. So, total outcomes about the first and the second letters= (4*4) = 16.
Therefore, following the same route, we can say, total possible outcomes about the first and the second letters = (4*4*4) = 64.
In how many ways 2 directors and 3 executives can be arranged for a meeting? If there are 6 chairs available two on one side and remaining four on the other side of the table and the two directors has to be together on one side and the executives on the other side.- a)24
- b)48
- c)720
- d)120
Correct answer is option 'B'. Can you explain this answer?
In how many ways 2 directors and 3 executives can be arranged for a meeting? If there are 6 chairs available two on one side and remaining four on the other side of the table and the two directors has to be together on one side and the executives on the other side.
a)
24
b)
48
c)
720
d)
120
|
Hansa Sharma answered |
3 executives in 4 chairs = 4P3 ways
Total no. of ways = 2! * 4P3
=> 2*1*(4!/1!)
= 2*24
= 48
Total no. of ways = 2! * 4P3
=> 2*1*(4!/1!)
= 2*24
= 48
A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is- a)164
- b)140
- c)112
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is
a)
164
b)
140
c)
112
d)
none of these
|
Neha Joshi answered |
Let us say that the two particular friends are A and B.
If A is invited among six guests and B is not, then: number of combinations to select 5 more guests from the remaining 8 friends:
C(8, 5) = 8 ! / (5! 3!) = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests = C(8, 5) = 56
Suppose both A and B are not included in the six guests list : then the number of such combinations = C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected = 140.
If A is invited among six guests and B is not, then: number of combinations to select 5 more guests from the remaining 8 friends:
C(8, 5) = 8 ! / (5! 3!) = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests = C(8, 5) = 56
Suppose both A and B are not included in the six guests list : then the number of such combinations = C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected = 140.
The number of ways in which 6 “ + “ and 4 “ – “ signs can be arranged in a line such that no two “ – “ signs occur together is- a)P(10,4)
- b)C(7,4)
- c)C(10,4)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The number of ways in which 6 “ + “ and 4 “ – “ signs can be arranged in a line such that no two “ – “ signs occur together is
a)
P(10,4)
b)
C(7,4)
c)
C(10,4)
d)
none of these
|
Sushil Kumar answered |
′+′ signs can be put in a row in one way creating seven gaps shown as arrows:
Now 4′−′ signs must be kept in these gaps. So, no tow ′−′ signs should be together.
Out of these 7 gaps 4 can be chosen in 7C4 ways.
Now 4′−′ signs must be kept in these gaps. So, no tow ′−′ signs should be together.
Out of these 7 gaps 4 can be chosen in 7C4 ways.
Number of signals that can be made using given 4 flags of which 3 are blue and 1 is red?- a)3i
- b)12
- c)4i
- d)4
Correct answer is option 'D'. Can you explain this answer?
Number of signals that can be made using given 4 flags of which 3 are blue and 1 is red?
a)
3i
b)
12
c)
4i
d)
4
|
Shiksha Academy answered |
Total flags = 4
Blue = 3
red = 1
Total no. of signals = (4!*3!)/3!
= 4
Blue = 3
red = 1
Total no. of signals = (4!*3!)/3!
= 4
A room has 8 doors. In how many ways, a man can enter in the room through one door and exit through a different door?- a)5040
- b)40320
- c)56
- d)8
Correct answer is option 'C'. Can you explain this answer?
A room has 8 doors. In how many ways, a man can enter in the room through one door and exit through a different door?
a)
5040
b)
40320
c)
56
d)
8
|
Naina Sharma answered |
The person has 8 options to enter the hall. For each of these 8 options, he has 7 options to exit the hall. Thus, he has 8 × 7 = 56 ways to enter and exit from different doors.
A coin is tossed 6 times, in how many throws can 4 heads and 2 tails be obtained?- a)24
- b)18
- c)15
- d)10
Correct answer is option 'C'. Can you explain this answer?
A coin is tossed 6 times, in how many throws can 4 heads and 2 tails be obtained?
a)
24
b)
18
c)
15
d)
10
|
|
Naina Bansal answered |
Whether we toss a coin 6 times or six coins one time the number of arrangement will remain same .
As to find number of ways we get 4 heads and 2 tails out of 6 times
The number of different ways in which a man can invite one or more of his 6 friends to dinner is- a)63
- b)15
- c)30
- d)120
Correct answer is option 'A'. Can you explain this answer?
The number of different ways in which a man can invite one or more of his 6 friends to dinner is
a)
63
b)
15
c)
30
d)
120
|
Om Desai answered |
He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.
1 friend can be selected out of 6 in 6C1 = 6 ways
2 friends can be selected out of 6 in 6C2 = 15 ways
3 friends can be selected out of 6 in 6C3 = 20 ways
4 friends can be selected out of 6 in 6C4 = 15 ways
5 friends can be selected out of 6 in 6C5 = 6 ways
6 friends can be selected out of 6 in 6C6 = 1 ways
Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63
1 friend can be selected out of 6 in 6C1 = 6 ways
2 friends can be selected out of 6 in 6C2 = 15 ways
3 friends can be selected out of 6 in 6C3 = 20 ways
4 friends can be selected out of 6 in 6C4 = 15 ways
5 friends can be selected out of 6 in 6C5 = 6 ways
6 friends can be selected out of 6 in 6C6 = 1 ways
Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63
The total number of ways of answering 5 objective questions, each question having four choices are- a)2880
- b)120
- c)1024
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
The total number of ways of answering 5 objective questions, each question having four choices are
a)
2880
b)
120
c)
1024
d)
None of the above
|
|
Preeti Iyer answered |
The total ways of answering would be. 4x4x4x4x4=4^5 that will be 1024.
In how many ways, a party of 5 men and 5 women be seated at a circular table, so that no two women are adjacent?- a)720
- b)14400
- c)2880
- d)1440
Correct answer is option 'C'. Can you explain this answer?
In how many ways, a party of 5 men and 5 women be seated at a circular table, so that no two women are adjacent?
a)
720
b)
14400
c)
2880
d)
1440
|
Ayush Joshi answered |
Lets first place the men (M). '*' here indicates the linker of round table
* M -M - M - M - M *
which is in (5-1)! ways
So we have to place the women in between the men which is on the 5 empty seats ( 4 -'s and 1 linker i.e * )
SO 5 women can sit on 5 seats in (5)! ways or
1st seat in 5 ways
2nd seat 4
3rd seat 3
4th seat 2
5th seat 1
i.e 5*4*3*2*1 ways
So the answer is 5! * 4! = 2880
In how many ways can 4 red, 3 yellow and 2 green chairs be arranged in a row if the chairs of the same colour are indistinguishable?- a)15120
- b)1260
- c)362880
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
In how many ways can 4 red, 3 yellow and 2 green chairs be arranged in a row if the chairs of the same colour are indistinguishable?
a)
15120
b)
1260
c)
362880
d)
None of the above
|
|
Hansa Sharma answered |
Total no of balls = 9
red balls = 4
yellow balls = 3
green balls = 2
Total no. of arrangements = 9!/(4!*3!*2!)
= 1260
red balls = 4
yellow balls = 3
green balls = 2
Total no. of arrangements = 9!/(4!*3!*2!)
= 1260
Among 7 flags 4 are of red colour and the rest are all different colours.How mant different signals can be generated using these flags?- a)240
- b)210
- c)180
- d)120
Correct answer is option 'B'. Can you explain this answer?
Among 7 flags 4 are of red colour and the rest are all different colours.How mant different signals can be generated using these flags?
a)
240
b)
210
c)
180
d)
120
|
Poonam Reddy answered |
Total flags = 4
Different colour = 3
red = 4
Total no. of signals = 7P3
= 7!/(4!)
= 210
Different colour = 3
red = 4
Total no. of signals = 7P3
= 7!/(4!)
= 210
Sonia has 10 balloons out of which 5 are red, 2 white, 2 blue and 1pink, which she wants to use for the decoration. Her favourite pink colour balloon should be filled with toffees and should be put at the centre of the room above the cake table and remaining 9 at the wall behind the cake table. How many ways she can arrange the balloons?- a)956
- b)756
- c)9!
- d)7560
Correct answer is option 'B'. Can you explain this answer?
Sonia has 10 balloons out of which 5 are red, 2 white, 2 blue and 1pink, which she wants to use for the decoration. Her favourite pink colour balloon should be filled with toffees and should be put at the centre of the room above the cake table and remaining 9 at the wall behind the cake table. How many ways she can arrange the balloons?
a)
956
b)
756
c)
9!
d)
7560
|
|
Pooja Shah answered |
1 balloon is fixed that is to be placed above the cake
Now she have to arrange 9 balloons = 9!/(5!*2!*2!)
(9*8*7*6*5!)/((5!*2*2)
= (9*8*7*6)/(2*2)
= 756
Now she have to arrange 9 balloons = 9!/(5!*2!*2!)
(9*8*7*6*5!)/((5!*2*2)
= (9*8*7*6)/(2*2)
= 756
A coin is tossed n times, the number of all the possible outcomes is- a)C(n , 2)
- b)P(n , 2)
- c)2n
- d)2n
Correct answer is option 'D'. Can you explain this answer?
A coin is tossed n times, the number of all the possible outcomes is
a)
C(n , 2)
b)
P(n , 2)
c)
2n
d)
2n
|
Krishna Iyer answered |
If coin is tossed n times then possible number of outcomes = 2n
- a)4/5!
- b)5/5!
- c)6/5!
- d)1/5!
Correct answer is option 'C'. Can you explain this answer?
a)
4/5!
b)
5/5!
c)
6/5!
d)
1/5!
|
|
Pooja Shah answered |
1/4! + 1/5!
= 1/4![1 + 1/5]
1/4![6/5]
=> 6/5!
= 1/4![1 + 1/5]
1/4![6/5]
=> 6/5!
The number of ways in which three different rings can be worn in four fingers with at most one in each finger, are- a)12
- b)64
- c)24
- d)6
Correct answer is 'C'. Can you explain this answer?
The number of ways in which three different rings can be worn in four fingers with at most one in each finger, are
a)
12
b)
64
c)
24
d)
6
|
Gaurav Kumar answered |
The total number of ways is same as the number of arrangements of 4 fingers, taken 3 at a time.
So, required number of ways = 4P3
= 4!/(4-3)!
= 4!/1!
= 4! => 24
So, required number of ways = 4P3
= 4!/(4-3)!
= 4!/1!
= 4! => 24
A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is- a)164
- b)140
- c)112
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is
a)
164
b)
140
c)
112
d)
none of these
|
|
Alok Mehta answered |
Case I : Both of them don't attend the party.then, no of ways = 8C6 = 28Case
II : Either of them is selected for party.then, no of ways = 2C1 * 8C5 = 112
Total no of ways = 112+28 = 140
If (n + 1)! = 20(n – 1)!, then n is equal to- a)20
- b)5
- c)-5
- d)4
Correct answer is option 'D'. Can you explain this answer?
If (n + 1)! = 20(n – 1)!, then n is equal to
a)
20
b)
5
c)
-5
d)
4
|
Vikas Kapoor answered |
(n + 1)! = 20 (n – 1)!
n (n + 1) = 20
(n – 4) (n + 5) = 0
Since, (n – 1)! exists, n ≥ 1
So, n = 4
n (n + 1) = 20
(n – 4) (n + 5) = 0
Since, (n – 1)! exists, n ≥ 1
So, n = 4
The figures 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is- a)98
- b)72
- c)90
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The figures 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is
a)
98
b)
72
c)
90
d)
none of these
|
|
Riya Banerjee answered |
There are in total 6 numbers, 4,5,6,7,8.
Now consider the number 56000
Consider the numbers of the form
56−−−.
Considering no repetitions we get
3×2×1 = 6 numbers.
Similarly for 57−−− and 58−−−.
Hence 3×6 = 18 numbers.
Now consider the numbers starting with 6.
6−−−−
We get 4×3×2×1 = 24.
Similarly for the numbers starting with 7 and 8 we get in total 24 numbers each.
Hence total number of numbers greater than 56000 will be
= (24×3)+18
= 72+18 = 90
Now consider the number 56000
Consider the numbers of the form
56−−−.
Considering no repetitions we get
3×2×1 = 6 numbers.
Similarly for 57−−− and 58−−−.
Hence 3×6 = 18 numbers.
Now consider the numbers starting with 6.
6−−−−
We get 4×3×2×1 = 24.
Similarly for the numbers starting with 7 and 8 we get in total 24 numbers each.
Hence total number of numbers greater than 56000 will be
= (24×3)+18
= 72+18 = 90
The number of all possible positive integral solutions of the equation xyz = 30 is- a)25
- b)26
- c)27
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The number of all possible positive integral solutions of the equation xyz = 30 is
a)
25
b)
26
c)
27
d)
none of these
|
|
Paridhi Nagori answered |
27 possible solutions
X-Y-Z
1-1-30 3 sets
1-2-15 6 sets
1-3-10 6 sets
1-5-6 6 sets
2-3-5 6 sets
The number of three digit numbers having atleast one digit as 5 is- a)225
- b)246
- c)648
- d)252
Correct answer is option 'D'. Can you explain this answer?
The number of three digit numbers having atleast one digit as 5 is
a)
225
b)
246
c)
648
d)
252
|
|
Ravi Sharma answered |
These digit number without digit 5 →100....999
→ these are 900 three-digit number
→ from 100 to 199 → 19 number with 5.
200−299→19
300−399→19
400−499→19
600−699→19
700−799→19
800−899→19
900−999→19
500−599→100
total number with 5=19×8+100 for (500-599)
=152+100
=252
→ these are 900 three-digit number
→ from 100 to 199 → 19 number with 5.
200−299→19
300−399→19
400−499→19
600−699→19
700−799→19
800−899→19
900−999→19
500−599→100
total number with 5=19×8+100 for (500-599)
=152+100
=252
The number of all three digit even numbers such that if 5 is one of the digits then next digit is 7 is- a)360
- b)370
- c)365
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The number of all three digit even numbers such that if 5 is one of the digits then next digit is 7 is
a)
360
b)
370
c)
365
d)
none of these
|
|
Raghav Bansal answered |
Correct Answer :- c
Explanation : As given condition is , 5 must be followed by 7. So only possible way is 57X where X denotes 0,2,4,6 and 8.
So,total no. of ways=1×1×5=5
Hence, total ways in which we can make a 3-digit even no. without violating given condition are:
360+5=365
Numbers greater than 1000 but not greater than 4000 are to be formed with the digits 0, 1, 2, 3, 4, allowing repetitions, the number of possible numbers is- a)370
- b)376
- c)365
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
Numbers greater than 1000 but not greater than 4000 are to be formed with the digits 0, 1, 2, 3, 4, allowing repetitions, the number of possible numbers is
a)
370
b)
376
c)
365
d)
none of these.
|
|
Niks Niks answered |
By reapeating digit 0,1,3,4 there are 376 numbers are formed between 1000- 4000
In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included? - a)3
- b)30
- c)64
- d)84
Correct answer is option 'C'. Can you explain this answer?
In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included?
a)
3
b)
30
c)
64
d)
84
|
|
Geetika Shah answered |
At least one yellow chair means “total way no yellow chair
Total ways to select 3 chairs form the total (2+3+4) chairs 9C3
9C3 = 9!/3!*6! = 84ways
Now, we dont want even one yellow chair
So, we should select 3 chairs from 6 chairs (2green & 4blue) = 6C3
6C3 = 6!/3!3! = 20 ways
hence , ways to select at least one yellow chair = 84-20 = 64 ways.
Total ways to select 3 chairs form the total (2+3+4) chairs 9C3
9C3 = 9!/3!*6! = 84ways
Now, we dont want even one yellow chair
So, we should select 3 chairs from 6 chairs (2green & 4blue) = 6C3
6C3 = 6!/3!3! = 20 ways
hence , ways to select at least one yellow chair = 84-20 = 64 ways.
An online examination have 12 question out of which we havethe alternative to select the answer. Choose how many ways can be there in which one can answer the question
- a)312+1
- b)212
- c)312−1
- d)312
Correct answer is option 'C'. Can you explain this answer?
An online examination have 12 question out of which we havethe alternative to select the answer. Choose how many ways can be there in which one can answer the question
a)
312+1
b)
212
c)
312−1
d)
312
|
Lavanya Menon answered |
No.of choices for each question = 3
There are six questions so total = (3)12
So no.of ways = (3)12 - 1
There are six questions so total = (3)12
So no.of ways = (3)12 - 1
On a railway track, there are 20 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other is- a)C(20,2)
- b)P(20,2)
- c)400
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
On a railway track, there are 20 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other is
a)
C(20,2)
b)
P(20,2)
c)
400
d)
none of these
|
Chirag Verma answered |
Number of tickets selected from first station =20
from second =19
.... for last station =0
We have to select 2 consecutive stations
so total number of possible tickets = P(20,2)
from second =19
.... for last station =0
We have to select 2 consecutive stations
so total number of possible tickets = P(20,2)
The number of permutations of n different objects taken r at a time, where repetition is allowed- a)nr-1
- b)nr
- c)n!
- d)(n-1)!
Correct answer is option 'B'. Can you explain this answer?
The number of permutations of n different objects taken r at a time, where repetition is allowed
a)
nr-1
b)
nr
c)
n!
d)
(n-1)!
|
Shreya Sengupta answered |
Understanding Permutations with Repetition
When dealing with permutations of n different objects taken r at a time with repetition allowed, we can derive the formula for the total arrangements possible.
Key Concepts
- Permutations: Arranging objects in a specific order.
- Repetition Allowed: Each object can be chosen multiple times.
Formula Derivation
1. Choice for Each Position:
- For each of the r positions in the arrangement, we can choose any of the n objects.
2. Total Arrangements:
- Since repetition is allowed, each of the r positions can independently select from n options.
- Thus, the total number of permutations can be calculated as:
- n choices for the first position
- n choices for the second position
- ...
- n choices for the r-th position
3. Mathematical Representation:
- This leads to the expression: n * n * n * ... (r times) = n^r.
Conclusion
Thus, the number of permutations of n different objects taken r at a time, where repetition is allowed, is given by the formula n^r.
Correct Option
The correct answer is option 'B', which represents this concept accurately as n^r.
When dealing with permutations of n different objects taken r at a time with repetition allowed, we can derive the formula for the total arrangements possible.
Key Concepts
- Permutations: Arranging objects in a specific order.
- Repetition Allowed: Each object can be chosen multiple times.
Formula Derivation
1. Choice for Each Position:
- For each of the r positions in the arrangement, we can choose any of the n objects.
2. Total Arrangements:
- Since repetition is allowed, each of the r positions can independently select from n options.
- Thus, the total number of permutations can be calculated as:
- n choices for the first position
- n choices for the second position
- ...
- n choices for the r-th position
3. Mathematical Representation:
- This leads to the expression: n * n * n * ... (r times) = n^r.
Conclusion
Thus, the number of permutations of n different objects taken r at a time, where repetition is allowed, is given by the formula n^r.
Correct Option
The correct answer is option 'B', which represents this concept accurately as n^r.
In a multiple choice question, there are 4 alternatives, of which one or more are correct. The number of ways in which a candidate can attempt this question is
- a)4
- b)25
- c)16
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
In a multiple choice question, there are 4 alternatives, of which one or more are correct. The number of ways in which a candidate can attempt this question is
a)
4
b)
25
c)
16
d)
none of these
|
|
Geetika Shah answered |
To solve this problem, we need to calculate the total number of ways a candidate can attempt the multiple-choice question where there are 4 alternatives, and one or more can be correct.
Step 1: Consider each alternative
Each of the 4 alternatives can either be:
- Selected (included in the answer)
- Not selected (excluded from the answer)
So, for each alternative, there are 2 choices (select or not select).
Step 2: Calculate the total number of combinations
If there were no restrictions (i.e., selecting none is allowed), the total number of combinations would be 24=16
Step 3: Subtract the invalid case
Since at least one alternative must be selected (one or more are correct), we subtract the one case where none of the alternatives are selected:
24−1=16−1=15
Find the number of words with or without meaning which can be made using all the letters of the word SWEET.- a)120
- b)24
- c)48
- d)60
Correct answer is option 'D'. Can you explain this answer?
Find the number of words with or without meaning which can be made using all the letters of the word SWEET.
a)
120
b)
24
c)
48
d)
60
|
Md Jisan answered |
Sweet has 5 letters, so it gives 5! but there are two letters which are same , so number of words will be = (5! ÷ 2!) = 60
How many three letter codes can be formed by only vowels of English alphabets, given that repetition of letters is allowed?- a)125
- b)5
- c)25
- d)5!
Correct answer is option 'A'. Can you explain this answer?
How many three letter codes can be formed by only vowels of English alphabets, given that repetition of letters is allowed?
a)
125
b)
5
c)
25
d)
5!
|
Advait Chakraborty answered |
Number of Three-Letter Codes with Repetition of Vowels
To determine the number of three-letter codes that can be formed using only vowels of the English alphabet, we need to consider the following:
Step 1: Identify the Vowels in the English Alphabet
The English alphabet consists of 26 letters, out of which five are vowels: A, E, I, O, and U. These are the letters we will be using to form the three-letter codes.
Step 2: Determine the Number of Options for Each Position
Since repetition of letters is allowed, we can choose any vowel for each position. Therefore, for each position, there are five options (A, E, I, O, or U).
Step 3: Multiply the Number of Options for Each Position
To find the total number of three-letter codes, we need to multiply the number of options for each position. Since there are three positions, we multiply 5 options by itself three times.
5 x 5 x 5 = 125
Therefore, the correct answer is option 'A', 125.
Explanation:
When forming three-letter codes using only vowels of the English alphabet, repetition is allowed. This means that each position can be filled with any vowel multiple times.
For the first position, there are five options: A, E, I, O, or U. Similarly, for the second and third positions, there are also five options each.
To find the total number of three-letter codes, we multiply the number of options for each position. Since there are three positions, we multiply 5 options by itself three times:
5 x 5 x 5 = 125
Therefore, there are 125 possible three-letter codes that can be formed using only vowels of the English alphabet when repetition is allowed.
To determine the number of three-letter codes that can be formed using only vowels of the English alphabet, we need to consider the following:
Step 1: Identify the Vowels in the English Alphabet
The English alphabet consists of 26 letters, out of which five are vowels: A, E, I, O, and U. These are the letters we will be using to form the three-letter codes.
Step 2: Determine the Number of Options for Each Position
Since repetition of letters is allowed, we can choose any vowel for each position. Therefore, for each position, there are five options (A, E, I, O, or U).
Step 3: Multiply the Number of Options for Each Position
To find the total number of three-letter codes, we need to multiply the number of options for each position. Since there are three positions, we multiply 5 options by itself three times.
5 x 5 x 5 = 125
Therefore, the correct answer is option 'A', 125.
Explanation:
When forming three-letter codes using only vowels of the English alphabet, repetition is allowed. This means that each position can be filled with any vowel multiple times.
For the first position, there are five options: A, E, I, O, or U. Similarly, for the second and third positions, there are also five options each.
To find the total number of three-letter codes, we multiply the number of options for each position. Since there are three positions, we multiply 5 options by itself three times:
5 x 5 x 5 = 125
Therefore, there are 125 possible three-letter codes that can be formed using only vowels of the English alphabet when repetition is allowed.
In a class 6 students have to be arranged for a photograph. If the prefect and the vice must occupy the positions at either ends, how many ways the students can be arranged?- a)58
- b)48
- c)64
- d)32
Correct answer is option 'B'. Can you explain this answer?
In a class 6 students have to be arranged for a photograph. If the prefect and the vice must occupy the positions at either ends, how many ways the students can be arranged?
a)
58
b)
48
c)
64
d)
32
|
|
Om Desai answered |
In a class of 6 students. Two students at both ends. Remaining four students have to be arranged
= 2 * 4!
= 2 * 24
= 48
= 2 * 4!
= 2 * 24
= 48
The number of even numbers that can be formed by using all the digits 1, 2, 3, 4, and 5 (without repetitions) is
- a)48
- b)162
- c)1250
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of even numbers that can be formed by using all the digits 1, 2, 3, 4, and 5 (without repetitions) is
a)
48
b)
162
c)
1250
d)
none of these
|
Gaurav Choudhury answered |
Solution:
To form an even number, the unit digit must be 2 or 4. Let's consider both cases:
Case 1: Unit digit is 2
In this case, we have to form a number using the digits 1, 3, 4, 5 as well, without repetition. We can choose the tens digit in 4 ways (since we cannot use 2), the hundreds digit in 3 ways, the thousands digit in 2 ways, and the ten-thousands digit in 1 way. So there are 4 × 3 × 2 × 1 = 24 even numbers that can be formed with 2 as the unit digit.
Case 2: Unit digit is 4
This is similar to the previous case, except we cannot use 4 as the thousands digit. So we can choose the tens digit in 4 ways, the hundreds digit in 3 ways, the thousands digit in 2 ways (using either 1 or 3), and the ten-thousands digit in 1 way. So there are 4 × 3 × 2 × 1 = 24 even numbers that can be formed with 4 as the unit digit.
Total number of even numbers = 24 + 24 = 48
Therefore, option (a) is the correct answer.
To form an even number, the unit digit must be 2 or 4. Let's consider both cases:
Case 1: Unit digit is 2
In this case, we have to form a number using the digits 1, 3, 4, 5 as well, without repetition. We can choose the tens digit in 4 ways (since we cannot use 2), the hundreds digit in 3 ways, the thousands digit in 2 ways, and the ten-thousands digit in 1 way. So there are 4 × 3 × 2 × 1 = 24 even numbers that can be formed with 2 as the unit digit.
Case 2: Unit digit is 4
This is similar to the previous case, except we cannot use 4 as the thousands digit. So we can choose the tens digit in 4 ways, the hundreds digit in 3 ways, the thousands digit in 2 ways (using either 1 or 3), and the ten-thousands digit in 1 way. So there are 4 × 3 × 2 × 1 = 24 even numbers that can be formed with 4 as the unit digit.
Total number of even numbers = 24 + 24 = 48
Therefore, option (a) is the correct answer.
The number of all numbers that can be formed by using some or all of the digits 1, 3, 5, 7, 9 (without repetitions) is- a)325
- b)120
- c)32
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of all numbers that can be formed by using some or all of the digits 1, 3, 5, 7, 9 (without repetitions) is
a)
325
b)
120
c)
32
d)
none of these
|
|
Pooja Shah answered |
Out of 1, 3, 5, 7, 9
No. of 1-digit numbers = 5
No. of 2-digit numbers = 5*4 = 20
No. of 3-digit numbers = 5*4*3 = 60
No. of 4-digit numbers = 5*4*3*2 = 120
No. of 5-digit numbers = 5*4*3*2*1 = 120
Total no. of numbers = 5 + 20 + 60 + 120 + 120 = 325
No. of 1-digit numbers = 5
No. of 2-digit numbers = 5*4 = 20
No. of 3-digit numbers = 5*4*3 = 60
No. of 4-digit numbers = 5*4*3*2 = 120
No. of 5-digit numbers = 5*4*3*2*1 = 120
Total no. of numbers = 5 + 20 + 60 + 120 + 120 = 325
In how many ways can a mixed doubles tennis game be arranged from a group of 10 players consisting of 6 men and 4 women- a)180
- b)90
- c)120
- d)48
Correct answer is option 'A'. Can you explain this answer?
In how many ways can a mixed doubles tennis game be arranged from a group of 10 players consisting of 6 men and 4 women
a)
180
b)
90
c)
120
d)
48
|
Anireddy Charan answered |
The number of ways a doubles tennis game be arranged = 45 ways (i.e.^10C2)
But in a mixed doubles tennis game, you will get 4 ways
i.e. two women vs two men, one woman and a man vs two men, one woman and a man vs two women, one woman and a man vs one woman and a man.
So, the total number of ways = 45 x 4 ways = 180
How many different words can be formed using the letters of the word BHARAT, which begin with B and end with T?- a)36
- b)16
- c)24
- d)12
Correct answer is option 'D'. Can you explain this answer?
How many different words can be formed using the letters of the word BHARAT, which begin with B and end with T?
a)
36
b)
16
c)
24
d)
12
|
Sarita Yadav answered |
If the first & last letter is fixed, then we find out, the number of permutations of the remaining letters, i.e. 4
= 4!/2!
= 4*3*2!/2!
= 12
= 4!/2!
= 4*3*2!/2!
= 12
How many ways can three white and three red balloons be arranged in a row?- a)6!
- b)10
- c)20
- d)3!.3!
Correct answer is option 'C'. Can you explain this answer?
How many ways can three white and three red balloons be arranged in a row?
a)
6!
b)
10
c)
20
d)
3!.3!
|
Sai Iyer answered |
Arranging Three White and Three Red Balloons in a Row
To determine the number of ways in which three white and three red balloons can be arranged in a row, we can use the formula for permutations of objects with repetition.
Formula: n! / (n1!n2!...nk!), where n is the total number of objects, and n1, n2,...nk are the number of objects of each type.
Using this formula, we can calculate the number of arrangements as follows:
n = 6 (total number of balloons)
n1 = 3 (number of white balloons)
n2 = 3 (number of red balloons)
Number of arrangements = 6! / (3! 3!) = 20
Therefore, there are 20 ways in which three white and three red balloons can be arranged in a row.
Explanation:
- The formula for permutations with repetition is n! / (n1!n2!...nk!), where n is the total number of objects, and n1, n2,...nk are the number of objects of each type.
- In this case, we have a total of six balloons, with three white and three red balloons.
- Using the formula, we can calculate the number of arrangements as 6! / (3! 3!) = 20.
- This means that there are 20 ways in which we can arrange the three white and three red balloons in a row.
- Therefore, the correct answer is option C.
To determine the number of ways in which three white and three red balloons can be arranged in a row, we can use the formula for permutations of objects with repetition.
Formula: n! / (n1!n2!...nk!), where n is the total number of objects, and n1, n2,...nk are the number of objects of each type.
Using this formula, we can calculate the number of arrangements as follows:
n = 6 (total number of balloons)
n1 = 3 (number of white balloons)
n2 = 3 (number of red balloons)
Number of arrangements = 6! / (3! 3!) = 20
Therefore, there are 20 ways in which three white and three red balloons can be arranged in a row.
Explanation:
- The formula for permutations with repetition is n! / (n1!n2!...nk!), where n is the total number of objects, and n1, n2,...nk are the number of objects of each type.
- In this case, we have a total of six balloons, with three white and three red balloons.
- Using the formula, we can calculate the number of arrangements as 6! / (3! 3!) = 20.
- This means that there are 20 ways in which we can arrange the three white and three red balloons in a row.
- Therefore, the correct answer is option C.
A team of 7 players is to be formed out of 5 under 19 players and 6 senior players. In how many ways, the team can be chosen when at least 4 senior players are included?- a)185
- b)215
- c)115
- d)125
Correct answer is option 'B'. Can you explain this answer?
A team of 7 players is to be formed out of 5 under 19 players and 6 senior players. In how many ways, the team can be chosen when at least 4 senior players are included?
a)
185
b)
215
c)
115
d)
125
|
|
Om Desai answered |
No. of ways to select 4 senior and 3 U-19 players = 6C4 * 5C3 = 150
No. of ways to select 5 senior and 2 U-19 players = 6C5 * 5C2 = 60
No. of ways to select 6 senior and 1 U-19 players = 6C6 * 5C1 = 5
Total no. of ways to select the team = 150 + 60 + 5 = 215
No. of ways to select 5 senior and 2 U-19 players = 6C5 * 5C2 = 60
No. of ways to select 6 senior and 1 U-19 players = 6C6 * 5C1 = 5
Total no. of ways to select the team = 150 + 60 + 5 = 215
5 boys and 5 girls are to be seated around a table such that boys and girls sit alternately. The number of ways of seating them is- a)5! × 4!
- b)4! × 4!
- c)5! × 5!
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
5 boys and 5 girls are to be seated around a table such that boys and girls sit alternately. The number of ways of seating them is
a)
5! × 4!
b)
4! × 4!
c)
5! × 5!
d)
none of these
|
|
Lavanya Menon answered |
First we fix the alternate position of the girls. Five girls can be seated around the circle in (5−1)!=4! , 5 boys can be seated in five -vacant place by 5!
∴ Required number of ways =4!×5!
∴ Required number of ways =4!×5!
Number of ways in which 10 different things can be divided into two groups containing 6 and 4 things respectively is- a)P(10,2)
- b)P(10,4)
- c)C(10,4)
- d)P(10,6)
Correct answer is option 'C'. Can you explain this answer?
Number of ways in which 10 different things can be divided into two groups containing 6 and 4 things respectively is
a)
P(10,2)
b)
P(10,4)
c)
C(10,4)
d)
P(10,6)
|
|
Ananya Das answered |
Forming the first group by choosing 4 things out of 10 things,the total number of ways will be =10C4
Now,forming these group by choosing 6 things,the total number of ways =
6C6
Therefore,the total number of ways = 10C4∗6C6
= C(10,4)
Now,forming these group by choosing 6 things,the total number of ways =
6C6
Therefore,the total number of ways = 10C4∗6C6
= C(10,4)
The number of arrangements of n different things taken r at a time which include a particular thing is- a)P(n-1, r-1)
- b)nP(n-1,r)
- c)rP(n-1,r-1)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The number of arrangements of n different things taken r at a time which include a particular thing is
a)
P(n-1, r-1)
b)
nP(n-1,r)
c)
rP(n-1,r-1)
d)
none of these
|
Geetika Shah answered |
Total Permutations of n different things taken r at a time when a particular item is always included in the arrangement is : r * (n-1 p r-1).
The number of triangles that can be formed with 6 points on a circle is- a)15
- b)6
- c)20
- d)12
Correct answer is option 'C'. Can you explain this answer?
The number of triangles that can be formed with 6 points on a circle is
a)
15
b)
6
c)
20
d)
12
|
Lakshmi Sharma answered |
Understanding the Problem
To determine how many triangles can be formed from 6 points on a circle, we need to consider the properties of combinations.
Combination Concept
- A triangle is formed by selecting 3 points.
- The formula to calculate combinations is given by C(n, r) = n! / (r! * (n - r)!), where n is the total number of points, and r is the number of points to choose.
Calculating Combinations
- Here, n = 6 (the points on the circle) and r = 3 (points needed to form a triangle).
- We apply the combination formula:
C(6, 3) = 6! / (3! * (6 - 3)!)
= 6! / (3! * 3!)
= (6 × 5 × 4) / (3 × 2 × 1)
= 20.
Conclusion
- Therefore, the number of triangles that can be formed with 6 points on a circle is 20.
- The correct answer is option 'C'.
This approach highlights that any three points chosen from the six points on the circle will always form a triangle, as no three points are collinear on a circle.
To determine how many triangles can be formed from 6 points on a circle, we need to consider the properties of combinations.
Combination Concept
- A triangle is formed by selecting 3 points.
- The formula to calculate combinations is given by C(n, r) = n! / (r! * (n - r)!), where n is the total number of points, and r is the number of points to choose.
Calculating Combinations
- Here, n = 6 (the points on the circle) and r = 3 (points needed to form a triangle).
- We apply the combination formula:
C(6, 3) = 6! / (3! * (6 - 3)!)
= 6! / (3! * 3!)
= (6 × 5 × 4) / (3 × 2 × 1)
= 20.
Conclusion
- Therefore, the number of triangles that can be formed with 6 points on a circle is 20.
- The correct answer is option 'C'.
This approach highlights that any three points chosen from the six points on the circle will always form a triangle, as no three points are collinear on a circle.
Find Rank of word ‘wife ‘among the words that can be formed with its letters and arranged as in dictionary is- a)8
- b)24
- c)4
- d)18
Correct answer is option 'B'. Can you explain this answer?
Find Rank of word ‘wife ‘among the words that can be formed with its letters and arranged as in dictionary is
a)
8
b)
24
c)
4
d)
18
|
Aditya Sengupta answered |
♦Rank of SPAIN:
1.Alphabetic order of letters :- A,I,N,P,S
2.words starting with A: 4! = 24
3.words starting with I: 4! = 24
4.words starting with N :4! = 24
5.words starting with P :4! = 24
6.words starting with S :4! = 24
(words starting with SA :3!= 6
words starting with SI :3! = 6
words starting with SN :3! = 6
words starting with SP :3! = 6
{In words starting with SP first word is SPAIN} )
So now add the no. of words be4 SPAIN+1
Rank of SPAIN =24+24+24+24+6+6+6+1=115
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