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Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are [2002]
Required number of numbers = 5 x 6 x 6 x 4 = 36 x 20 = 720.
Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is [2002]
Required number of numbers = 3 x 5 x 5 x 5 = 375
Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are [2002]
We know that a number is divisible by 3 only when the sum of the digits is divisible by 3. The given digits are 0, 1, 2, 3, 4, 5.
Here the possible number of combinations of 5 digits out of 6 are ^{5}C_{4} = 5, which are as follows
– 1 + 2 + 3 + 4 + 5 = 15 = 3 × 5
0 + 2 + 3 + 4 + 5 = 14 (not divisible by 3)
0 + 1 + 3 + 4 + 5 = 13 (not divisible by 3)
0 + 1 + 2 + 4 + 5 = 12 = 3 × 4
0 + 1 + 2 + 3 + 5 = 11 (not divisible by 3)
0 + 1 + 2 + 3 + 4 = 10 ( not divisible by 3)
Thus the number should contain the digits 1, 2, 3, 4, 5 or the digits 0, 1, 2, 4, 5.
Taking 1, 2, 3, 4, 5, the 5 digit numbers are = 5! = 120
Taking 0, 1, 2, 4, 5, the 5 digit numbers are = 5! – 4! = 96
∴ Total number of numbers = 120 + 96 = 216
The sum of integers from 1 to 100 that are divisible by 2 or 5 is[2002]
Required sum = (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... + 100)
– (10 + 20 + ... + 100)
= 2550 + 1050 – 530 = 3050.
If ^{n}C_{r} denotes the number of combination of n things taken r at a time, then the expression ^{n}C_{r +1} + ^{n}C_{ r 1 }+ 2 x ^{n}C_{r} equals[2003]
^{n }C_{r +1 }+ ^{n }C_{r 1}+ 2^{n}C_{r} = ^{n}C_{r 1}+^{ n }C_{r} + ^{n}C_{r }+ ^{n}C_{r +1}
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is[2003]
As for given question two cases are possible.
(i) Selecting 4 out of first five question and 6 out of remaining 8 question
= ^{5}C_{4} x ^{8 }C_{6} = 140 choices.
(ii) Selecting 5 out of first five question and 5 out of remaining 8 questions
= ^{5}C_{5 }x ^{8} C_{5} = 56 choices.
∴ total number of choices =140 + 56 = 196.
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by[2003]
No. of ways in which 6 men can be arranged at a round table = (6  1)! = 5!
Now women can be arranged in ^{6 }P_{5} = 6! Ways.
Total Number of ways = 6! × 5!
How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order [2004]
Total number of arran gemen ts of letters in the word GARDEN = 6 ! = 720 there are two vowels A and E, in half of the arrangements A preceeds E and other half A follows E.
So, vowels in alphabetical order in x 720 = 360
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is [2004]
We know that the number of ways of distributing n identical items among r persons, when each one of them receives at least one item is ^{n 1}C_{r 1}
∴ The required number of ways
If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number [2005]
Alphabetical order is A, C, H, I, N, S
No. of words starting with A – 5!
No. of words starting with C – 5!
No. of words starting with H – 5!
No. of words starting with I – 5!
No. of words starting with N – 5!
SACHIN – 1
∴ sachin appears at serial no 601
At an election , a voter may vote for an y number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is [2006]
^{10}C_{1} + ^{10}C_{2 }+ ^{10}C_{3} + ^{10}C_{4}
= 10 + 45 + 120 + 210 = 385
The set S = {1, 2, 3, ......., 12} is to be partitioned into three sets A , B, C of equal size. Thus A ∪ B ∪ C = S , A∩B =B∩C= A∩C= φ. The number of ways to partition S is [2007]
Set S = {1, 2, 3, ...... 12}
A∪B∪C = S, A∩B =B∩C=A∩C= φ.
∴ The number of ways to partition
= ^{12}C_{4 }× ^{8}C_{4} × ^{4}C_{4}
In a shop there are five types of icecreams available. A child buys six icecreams.Statement1 : The number of different ways the child can buy the six icecreams is ^{10}C_{5}.Statement 2 : The number of different ways the child can buy the six icecreams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. [2008]
The given situation in statement 1 is equivalent to find the non negative integral solutions of the equation x_{1} + x_{2} + x_{3} + x_{4} + x_{5 }= 6
which is coeff. of x^{6} in the expansion of (1 + x + x^{2} + x^{3} + .....∞)^{5}
= coeff. of x^{6} in (1– x)^{–5}
= coeff. of x6 in 1 + 5x +
∴ Statement 1 is wrong.
Number of ways of arranging 6A’s and 4B’s in a row
which is same as the number of ways the child can buy six icecreams.
∴ Statement 2 is true.
How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?
First let us arrange M, I, I, I, I, P, P Which can be done in
ways
√ M √ I √ I √ I √ I √ P √ P √
Now 4 S can be kept at any of the ticked places in ^{8}C_{4} ways so that no two S are adjacent.
Total required ways
From 6 different novels and 3 different dictionaries,4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is: [2009]
4 novels, out of 6 novels and 1 dictionary out of 3 can be selected in ^{6 }C_{4} x ^{3}C_{1} ways
Then 4 novels with one dictionary in the middle can be arranged in 4! ways.
∴ Total ways of arrangement = ^{6}C_{4} x ^{3}C_{1}x 4! = 1080
There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is [2010]
Total number of ways = ^{3}C_{2} x ^{9}C_{2}
= 3 x 36 = 108
Statement1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is ^{9}C_{3} .Statement2: The number of ways of choosing any 3 places from 9 different places is ^{9}C_{3} . [2011]
The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box empty is same as the number of ways of selecting (r – 1) places out of (n – 1) different places, that is ^{n1}C_{r 1 }.
Hence required number of ways = ^{101}C_{41} = ^{9}C_{3}
∴ Both statements are correct and second statement is a correct explanation of statement 1.
These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:[2012]
Number of required triangles
= ^{10}C_{3} – ^{6}C_{3}
= 120  20 =100
Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is : [2012]
Number of white balls = 10
Number of green balls = 9
an d Number of black balls = 7
∴ Required probability = (10 + 1) (9 + 1) ( 7 + 1) – 1 = 11.10.8 –1 = 879
[∵ The total number of ways of selecting one or more items from p identical items of one kind, q identical items of second kind; r identical items of third kind is ( p + 1) (q + 1) (r + 1) –1 ]
Let T_{n} be the number of all possible triangles formed by joining vertices of an nsided regular polygon. If T_{n+1} – T_{n} = 10, then the value of n is : [JEE M 2013]
We know, T_{n }=^{ n}C_{3}, T_{n+1} = ^{n+1}C_{3 }
ATQ, T_{n+1} – T_{n} = ^{n+1}C_{3} – ^{n}C_{3} = 10
⇒ ^{n}C_{2} = 10
⇒ n = 5.
The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is : [JEE M 2015]
Four digits number can be arranged in 3 × 4! ways.
Five digits number can be arranged in 5! ways.
Number of integers = 3 × 4! + 5! = 192.
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is : [JEE M 2016]
ALLMS
No. of words starting with
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