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Permutations & Combinations Questions and Answers

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91.

On a railway line there are 20 stops. A ticket is needed to travel between any 2 stops. How many different tickets would the government need to prepare to cater to all possibilities?

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.

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92.

How many four digit numbers can be formed using the digits {1, 3, 4, 5, 7,9}(repetition of digits is not allowed)?

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.

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93.

There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?

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.

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94.

There are 30 people in a group. If all shake hands with one another , how many handshakes are possible?

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.

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95.

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

Answer: C

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

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

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

\[ 84 - 20 = 64 \]

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96.

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

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.

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97.

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

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.

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98.

There are 8 routes from London to Delhi. And there are 6 routes from Delhi to Tokyo. In how many different ways can Raj travel from London to Tokyo via Delhi?

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.

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99.

A question paper had 10 questions. Each question could only be answered as true(T) of False(F). Each candidate answered all the questions, Yet no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible?

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.

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100.

In a class, there are 15 students. During a Christmas party all of them shook hands with each other only once. How many handshakes took place in the class?

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.

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