Test: Permutations And Combinations - 1 - Commerce MCQ

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

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.

Mixed doubles includes 2 men and 2 women.
Since 2 men are selected of 6 men
⇒ number of ways = ⁶C₂
Also 2 women are selected of 4 women
⇒ number of ways = ⁴C₂
But they also can be interchanged in 2 ways
∴ Total no. of ways 

One digit positive integers are 0,1,2,3,4 =5 positive integers
Number of one digit no. can be formed = 4
Number of two digit no. can be formed = 4*4 = 16
Number of three digit no. can be formed = 4*4*3 = 48
Number of  four digit no. can be formed = 4*4*3*2 = 96
Number of five digit no. can be formed = 4*4*3*2*1 = 96
Total = 4+ 16+ 48+ 96+ 96 = 260

The smallest number in the series is 1000, a 4-digit number.  

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left-most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

So, The left-most digits have values 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

So, the next 3 digits have values 5.

Hence, there are numbers.

or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

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

If coin is tossed n times then possible number of outcomes = 2^n​

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

 N!/(n-5)! = 60×(n-1)!/(n-1-3)!
n!/(n-5)! = 60×(n-1)!/(n-4)!
n(n-1)!/(n-5)!=60×(n-1)!/(n-4)×(n-5)!
n=60/(n-4)
n(n-4)=60
n^2-4n-60=0
(n-10)(n+6)=0
n=10 and n is not equal to -6.

Number of sides of a cube = 6
Number of colour of a cube = 6
Number of ways of colouring =?
Consider a cube in 3D space 
Each of the sides towards East West North South and remaning two up and down 
Now fixing up positions down is also fixed 
Total number of ways 
= 6!/(4×6)
​= 5×3×2 
= 30 Ways
Hence, the correct answer is 30 Ways

 The correct answer is A

Given are four faces and four different colours

So , number of colours for first face=4

No. of colours for 2nd face=3 (as one will be used by the first face)

No. of colours for 3rd face=2

No. of colours for 4th face=1

So total options are=4*3*2*1=4!=24

No of word formed taking E at first place = 3! 
Taking F at first place = 3! 
Taking I in the first place = 3!
W in first and E in second = 2!
W in first F in second and E in third = 1! 
W in first I in second and E in third = 1 
W in first I in second and F in third = 1 
W in first I in second and F in third and E in fourth = 1 
Total =24

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

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:

1. Selected (included in the answer)
2. 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

ANSWER :- c

Solution :- Number of such triangle = 6C3

= 6!/3!3!

=20

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)

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).

Beside the ground floor, there are seven floors.
The total number of ways in which each of the five persons can leave cabin at any of the 7 floors =7⁵
And the favorable number of ways, that is, the number of ways, in which 5 persons leave at different floors is ⁷P₅

No of ways to make a necklace out of 11 beads = (11-1)!/2 
= 10!/2
Out of 11 beads there are identical red and black heads.
No. of arrangements = (10)!/[2*(5!6!)]