Permutations & Combinations Questions and Answers

Answer: D

Each of the three digits can be any of 0 to 9, so the digits are independent.

\[ 10 \times 10 \times 10 = 1000 \text{ combinations} \]

\[ 1000 \times 6 = 6000 \text{ seconds} = \frac{6000}{60} = 100 \text{ minutes} \]

Repetition is allowed here - nothing stops a code such as 777 - so it is \( 10^{3} \) rather than \( 10 \times 9 \times 8 \).

Answer: B

The outward journey has 7 options; the return must avoid whichever was used, leaving 6.

\[ 7 \times 6 = 42 \]

Without the restriction it would be \( 7 \times 7 = 49 \) - the first option, and the difference is exactly the 7 forbidden repeats.

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: A

The two cases are mutually exclusive, so their counts add.

3 teachers and 3 boys:

\[ {}^{5}C_{3} \times {}^{4}C_{3} = 10 \times 4 = 40 \]

2 girls and 4 teachers:

\[ {}^{3}C_{2} \times {}^{5}C_{4} = 3 \times 5 = 15 \]

\[ 40 + 15 = 55 \]

Both cases give exactly six members, as required, and no group satisfies both conditions at once - so nothing is double-counted.

Answer: C

MARKER has six letters with R twice, and its vowels are A and E.

\[ \text{total arrangements} = \frac{6!}{2!} = 360 \]

Treating the two vowels as one block gives five units:

\[ \frac{5!}{2!} \times 2! = 60 \times 2 = 120 \text{ with the vowels together} \]

\[ 360 - 120 = 240 \]

The division by \( 2! \) throughout accounts for the identical R's.

Answer: D

The first and last digits are fixed at 4 and 7, so only the four middle positions are free - and they must avoid 4 and 7, leaving eight digits.

\[ 8 \times 7 \times 6 \times 5 = 1680 \]

Repetition is forbidden, so each position has one fewer choice than the last.

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: 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: 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: B

(i) Two particular members must be included. They take two of the four places, so two more come from the remaining six:

\[ {}^{6}C_{2} = 15 \]

(ii) Two particular members must not be included. All four come from the remaining six:

\[ {}^{6}C_{4} = 15 \]

The two answers coincide because \( {}^{6}C_{2} = {}^{6}C_{4} \) - choosing two to include is the same as choosing four to leave out.