Permutation and Combination Questions for CAT with Answers PDF

There are 26 English alphabet and we need to find three-lettered passwords in which three letters are different, so
The number of passwords with different letters = 26 × 25 × 24

otal number of outcomes = 6 × 6 × 6 = 216 … (1)
Number of outcomes in which no die shows 4 = 5 × 5 × 5 = 125 … (2)
Number of outcomes in which at least 1 die shows 4 = (1) - (2)
= 216 - 125 = 91

The hall can be illuminated by switching on at least one of the 10 lamps.
Therefore, the required number of ways is 2¹⁰ - 1 = 1023.
Hence, option (2) is the correct answer.

 Since we require at least 1 lady on the committee, we could have 1, 2, 3 or 4 ladies on the committee. A committee of 5 from among 10 people can be chosen in ¹⁰C₅ ways = 252 ways. These ways include committees where there are no ladies. A committee with no ladies on it will have 5 men from among 6, i.e. ⁶C₅ ways = 6 ways. So, the required number of ways = 252 - 6 = 246

We just have to select 5 digits from the 10 available digits. When these five digits are selected, there is only one way in which these can exist in decreasing order.
Number of ways to select 5 digits out of 10 digits = ¹⁰C₅

Distinct letters = C, O, L, E, G
Number of distinct 4-letter words = ⁵P₄ = 5 × 4 × 3 × 2 = 120

Formula for number of circular arrangements of n different things taken r = nPr/r
Here, n = 10, r = 3
Total number of arrangements

= 720/3
= 240

Number of ways for going from ground floor to 1st floor = 4
Number of ways for going from 1st to 2nd floor = 2
Number of ways for going from 2nd to 1st floor = 3
Number of ways for going from 1st to ground floor = 3
∴ Total number of ways = 4 × 2 × 3 × 3

6 men can be chosen from 10 men in ¹⁰C₆ ways.
5 women can be chosen from 12 women in ¹²C₅ ways.
So, 6 men and 5 women can be chosen in ¹⁰C₆ × ¹²C₅ ways.