Permutations & Combinations Questions and Answers

Answer: B

CORPORATION has eleven letters, but repetition of letters is not allowed in the word formed, so only the distinct letters may be used:

\[ C,\ O,\ R,\ P,\ A,\ T,\ I,\ N \rightarrow 8 \text{ distinct} \]

Order matters in a word, so this is a permutation:

\[ {}^{8}P_{3} = 8 \times 7 \times 6 = 336 \]

The repeated O's and R's contribute nothing extra, since no letter may be used twice.

Answer: C

The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

Answer: D

Numbers between 500 and 1000 are three-digit numbers starting with 5 or above. From the given digits, the first digit can be 5, 6 or 8.

\[ \text{first digit} = 3 \text{ choices} \]

The remaining two positions take any of the six unused digits, without repetition:

\[ 6 \times 5 = 30 \]

\[ 3 \times 30 = 90 \]

0 is available for the later positions but not the first, and 1000 itself is excluded by the range.

Answer: C

A ticket names a starting stop and a destination, so the two directions need different tickets: the order matters.

\[ {}^{20}P_{2} = 20 \times 19 = 380 \]

Treating the journey as unordered would give \( {}^{20}C_{2} = 190 \) - the number of station pairs, not of tickets.

Answer: A

Six digits are available and four positions are to be filled without repetition.

\[ {}^{6}P_{4} = 6 \times 5 \times 4 \times 3 = 360 \]

None of the digits is 0, so there is no restriction on the leading position.

Answer: D

Eighteen stations lie between the two named ones, so there are 20 stations in all.

A ticket names both a source and a destination, so order matters:

\[ {}^{20}P_{2} = 20 \times 19 = 380 \]

The 190 on offer counts station pairs rather than tickets, missing the return direction.

Answer: B

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

\[ {}^{30}C_{2} = \frac{30 \times 29}{2} = 435 \]

The 870 on offer is \( 30 \times 29 \), which counts each handshake twice - once from each participant's side.

Answer: C

The box holds \( 2 + 3 + 4 = 9 \) balls, six of them not black.

\[ \text{total} = {}^{9}C_{3} = 84 \]

\[ \text{no black} = {}^{6}C_{3} = 20 \]

\[ 84 - 20 = 64 \]

Answer: D

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

 Required number of numbers = (1 x 5 x 4) = 20.

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.