Permutations & Combinations Questions and Answers

Answer: B

A rectangle on the board is fixed by choosing two of the nine vertical lines and two of the nine horizontal ones.

\[ {}^{9}C_{2} \times {}^{9}C_{2} = 36 \times 36 = 1296 \]

The squares are counted by size: an \( n \times n \) square can sit in \( (9-n)^{2} \) positions.

\[ 8^{2} + 7^{2} + \cdots + 1^{2} = 204 \]

\[ \text{rectangles that are not squares} = 1296 - 204 = 1092 \]

So the answer is 1092 and 204. Every square is a rectangle, which is why the two counts must be separated before comparing.

Answer: B

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

 Required number of ways = (120 x 6) = 720.

Answer: D

The two choices are independent, so they multiply.

\[ 9 \times 12 = 108 \]

Each of the nine trousers can be paired with each of the twelve shirts. Adding them would answer a different question - how many garments he could pick one of.

Answer: B

The box holds \( 4 + 3 + 6 = 13 \) marbles.

\[ \text{total ways} = {}^{13}C_{2} = 78 \]

\[ \text{same colour} = {}^{4}C_{2} + {}^{3}C_{2} + {}^{6}C_{2} = 6 + 3 + 15 = 24 \]

\[ P = \frac{24}{78} \]

The three colours are mutually exclusive, so their counts add before dividing.

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.