Permutations and Combinations: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 2)

One particular person must always be included in the committee. That person’s seat is already decided, so we don’t choose for them. We only need to choose the remaining 2 members from the other 9 people. Number of ways = 9C2 = 36.

One particular person must always be included in the committee. That person’s seat is already decided, so we don’t choose for them. We only need to choose the remaining 2 members from the other 11 people. Number of ways = 11C2 = 55.

One particular person must always be excluded from the committee. That removes 1 person from the pool, leaving 12 people to choose from. We choose all 5 committee members from these remaining 12 people. Number of ways = 12C5 = 792.

One particular person must always be included in the committee. That person’s seat is already decided, so we don’t choose for them. We only need to choose the remaining 5 members from the other 7 people. Number of ways = 7C5 = 21.

One particular person must always be included in the committee. That person’s seat is already decided, so we don’t choose for them. We only need to choose the remaining 6 members from the other 9 people. Number of ways = 9C6 = 84.

One particular person must always be excluded from the committee. That removes 1 person from the pool, leaving 7 people to choose from. We choose all 5 committee members from these remaining 7 people. Number of ways = 7C5 = 21.

One particular person must always be excluded from the committee. That removes 1 person from the pool, leaving 16 people to choose from. We choose all 5 committee members from these remaining 16 people. Number of ways = 16C5 = 4368.

One particular person must always be included in the committee. That person’s seat is already decided, so we don’t choose for them. We only need to choose the remaining 4 members from the other 19 people. Number of ways = 19C4 = 3876.

We’re choosing 6 people out of 15, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 15C6 = 15! / (6! x 9!) = 5005.

We’re choosing 3 people out of 8, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 8C3 = 8! / (3! x 5!) = 56.

We’re choosing 3 people out of 17, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 17C3 = 17! / (3! x 14!) = 680.

We’re choosing 5 people out of 11, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 11C5 = 11! / (5! x 6!) = 462.

We’re choosing 5 people out of 20, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 20C5 = 20! / (5! x 15!) = 15504.

We’re choosing 6 people out of 17, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 17C6 = 17! / (6! x 11!) = 12376.

We’re choosing 4 people out of 20, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 20C4 = 20! / (4! x 16!) = 4845.

We’re choosing 3 people out of 19, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 19C3 = 19! / (3! x 16!) = 969.

We’re choosing 5 people out of 10, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 10C5 = 10! / (5! x 5!) = 252.

We’re choosing 2 people out of 11, where order does NOT matter. A committee is a group, not a sequence, so we use combinations. Number of ways = 11C2 = 11! / (2! x 9!) = 55.

Total arrangements of ‘DIRECTOR’ = 20160, from 8! adjusted for repeated letters. Arrangements where the vowels stay together = 2160, found using the block method. Every arrangement either has the vowels together or it does not. There is no other possibility. So arrangements with the vowels never together = total minus together. 20160 – 2160 = 18000.

Total arrangements of ‘MEASURE’ = 2520, from 7! adjusted for repeated letters. Arrangements where the vowels stay together = 288, found using the block method. Every arrangement either has the vowels together or it does not. There is no other possibility. So arrangements with the vowels never together = total minus together. 2520 – 288 = 2232.

Total arrangements of ‘COUSIN’ = 720, from 6! adjusted for repeated letters. Arrangements where the vowels stay together = 144, found using the block method. Every arrangement either has the vowels together or it does not. There is no other possibility. So arrangements with the vowels never together = total minus together. 720 – 144 = 576.

Total arrangements of ‘DESIGN’ = 720, from 6! adjusted for repeated letters. Arrangements where the vowels stay together = 240, found using the block method. Every arrangement either has the vowels together or it does not. There is no other possibility. So arrangements with the vowels never together = total minus together. 720 – 240 = 480.

Total arrangements of ‘GARDEN’ = 720, from 6! adjusted for repeated letters. Arrangements where the vowels stay together = 240, found using the block method. Every arrangement either has the vowels together or it does not. There is no other possibility. So arrangements with the vowels never together = total minus together. 720 – 240 = 480.

Total arrangements of ‘DAUGHTER’ = 40320, from 8! adjusted for repeated letters. Arrangements where the vowels stay together = 4320, found using the block method. Every arrangement either has the vowels together or it does not. There is no other possibility. So arrangements with the vowels never together = total minus together. 40320 – 4320 = 36000.

Treat the 3 vowels (IEO) of ‘DIRECTOR’ as a single block, since they must stay together. ‘DIRECTOR’ then reduces to 5 consonants plus 1 vowel-block, or 6 units to arrange. These 6 units can be arranged in 360 ways. Within the block, the 3 vowels can be arranged among themselves in 6 ways. By the counting principle, total arrangements = 360 x 6 = 2160.