Permutations & Combinations Questions and Answers

Answer: D

CORPORATION has eleven letters - O three times, R twice - with vowels O, O, A, I, O and consonants C, R, P, R, T, N.

Treating the five vowels as one block gives seven units, among which R repeats:

\[ \frac{7!}{2!} = 2520 \]

Within the block, the five vowels include three O's:

\[ \frac{5!}{3!} = 20 \]

\[ 2520 \times 20 = 50{,}400 \]

Both repetitions have to be divided out - the R's among the units and the O's inside the block.

Answer: B

Each leg is chosen independently, so the counts multiply.

\[ 8 \times 6 = 48 \]

Every one of the 8 routes to Delhi can be paired with every one of the 6 onward routes - the fundamental counting principle.

Answer: D

Each of the ten questions is answered independently as T or F.

\[ 2^{10} = 1024 \]

Since no two candidates matched, at most 1024 candidates could sit the paper. The 512 on offer is \( 2^{9} \), one question short.

Answer: C

A handshake involves two people and the order does not matter, so the count is a combination.

\[ {}^{15}C_{2} = \frac{15 \times 14}{2} = 105 \]

The 210 on offer is \( 15 \times 14 \), counting each handshake from both sides.

Answer: C

DETAIL has six distinct letters: vowels E, A, I and consonants D, T, L.

The odd positions are 1, 3 and 5 - three places for three vowels.

\[ \text{vowels} = 3! = 6 \]

\[ \text{consonants in the even positions} = 3! = 6 \]

\[ 6 \times 6 = 36 \]

The two counts match exactly, which is what makes the restriction satisfiable at all.

Answer: A

There are ten animals in all. "At least one cow" is easiest through the complement.

\[ \text{total} = {}^{10}C_{4} = 210 \]

\[ \text{no cow} = {}^{4}C_{4} = 1 \]

\[ 210 - 1 = 209 \]

With only four buffalos there is exactly one all-buffalo selection, which is what makes the complement so small.

Answer: C

\[ {}^{4}C_{3} \times {}^{6}C_{2} \times {}^{5}C_{3} \]

\[ = 4 \times 15 \times 10 = 600 \]

Each group is chosen from its own pool, so the three counts are independent and multiply.

Answer: A

The leader is to be a school girl or a school boy, and no one is both, so the two counts simply add.

\[ 12 + 10 = 22 \]

"Or" between mutually exclusive choices means addition; it would be multiplication only if a girl and a boy were both being chosen.

Answer: C

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G). 

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB. 

Hence total number of ways by changing flag = 3+ 6 +6 = 15

Answer: C

MATHEMATICS has eleven letters, with M, A and T each appearing twice. Its vowels are A, E, A, I and its consonants M, T, H, M, T, C, S.

Treating the four vowels as one block gives eight units, among which M and T each repeat:

\[ \frac{8!}{2!\,2!} = \frac{40320}{4} = 10{,}080 \]

Within the block the four vowels include two A's:

\[ \frac{4!}{2!} = 12 \]

\[ 10{,}080 \times 12 = 120{,}960 \]

Three separate repetitions have to be divided out - the M's and T's among the units, and the A's inside the block.