Test: Permutations and Combinations - 3 - JEE 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

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​

 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.

 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

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

Each object can be put either in box B₁ (say) or in box B₂ ( say).
So, there are two choices for each of the n objects.
Therefore the number of choices for n distinct objects is 2 × 2 ×…..× 2 = 2ⁿ the second box being empty.
Thus, there are 2ⁿ − 2 ways in which neither box is empty.

Since we want 4 points for a parallelogram, two each on sets of parallel lines perpendicular to each other, the required number = ⁴C₂ × ³C₂ = 18