Permutations and Combinations Questions and Answers

Answer: C

There were 14 people initially.

With n people, everyone shakes hands with everyone once: \(\binom{n}{2}\) handshakes. After 5 leave, the newcomer shakes hands with the n − 5 still present.

\[ \begin{aligned} \frac{n(n - 1)}{2} + (n - 5) &= 100 \\ n = 14: \ 91 + 9 &= 100 \end{aligned} \]

Checking the neighbours shows 14 is the only fit: n = 13 gives 78 + 8 = 86 and n = 15 gives 105 + 10 = 115.

Answer: A

There are 32 such numbers.

With the digits 3, 4, 5 and 7, a 3-digit number exceeds 500 exactly when its first digit is 5 or 7 (the smallest such number, 533, is already above 500).

\[ 2 \times 4 \times 4 = 32 \]

Repetition is allowed, so the tens and units digits each have 4 choices. 16 counts only one of the two leading digits, and 12 is the count when repetition is not allowed (2 × 3 × 2).

Answer: E

\({}^{n}C_{5}\) = 3003.

Permutations and combinations differ by the 5! orderings of the chosen items:

\[ \begin{aligned} {}^{n}C_{5} &= \frac{{}^{n}P_{5}}{5!} \\ &= \frac{360360}{120} = 3003 \end{aligned} \]

Here n = 15, since \(15 \times 14 \times 13 \times 12 \times 11 = 360360\), and \({}^{15}C_{5} = 3003\).

The step works because each combination of 5 objects can be arranged in 5! = 120 orders, and each order is a different permutation.

Answer: B

The probability is \(\frac{14}{99}\).

There are 12 members, 8 of them not teachers. A committee with no teachers chooses all 4 from those 8.

\[ \begin{aligned} P &= \frac{\binom{8}{4}}{\binom{12}{4}} = \frac{70}{495} \\ &= \frac{14}{99} \end{aligned} \]

Dividing 70 and 495 by 5 gives the simplified fraction. The other options would need 65, 80, 100 or 125 favourable committees out of 495.

Answer: D

There are 30 ways.

The person makes three independent choices in sequence: an entrance (2 ways), one of the three lifts from that entrance (3 ways) and a floor (5 ways). By the multiplication principle:

\[ 2 \times 3 \times 5 = 30 \]

15 forgets the choice of entrance, 10 forgets the choice of lift, and 25 and 50 do not correspond to the stages of the choice.

Answer: B

\(r = 2\).

Reading the left side as \({}^{9}P_{5}\):

\[ \begin{aligned} {}^{9}P_{5} &= 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = 15120 \\ {}^{6}P_{r} &= \frac{15120}{504} = 30 \end{aligned} \]

Since \({}^{6}P_{2} = 6 \times 5 = 30\), \(r = 2\). The other values give \({}^{6}P_{1} = 6\), \({}^{6}P_{3} = 120\), \({}^{6}P_{4} = 360\) and \({}^{6}P_{5} = 720\).

Answer: E

The sum is 6660.

The number of 3-digit numbers with distinct digits from 1, 2, 3, 4 is \({}^{4}P_{3} = 24\). By symmetry each digit appears in each place (hundreds, tens, units) \(24 \div 4 = 6\) times.

\[ \begin{aligned} \text{sum of each place} &= 6(1 + 2 + 3 + 4) = 60 \\ \text{total} &= 60 \times (100 + 10 + 1) = 6660 \end{aligned} \]

The other options are not multiples of 111, which every such total must be, so they can be rejected without further work.

Answer: D

The number of ways is 1771.

Getting only green balls means choosing all three from the 23 green balls (the balls are treated as distinguishable when counting selections):

\[ \begin{aligned} {}^{23}C_{3} &= \frac{23 \times 22 \times 21}{3 \times 2 \times 1} \\ &= 1771 \end{aligned} \]

As a check, all selections number \({}^{24}C_{3} = 2024\), and those including the black ball number \({}^{23}C_{2} = 253\); \(2024 - 253 = 1771\). The nearby values 1763 to 1765 are arithmetic slips.

Answer: C

The value of \(n\) is 15.

The paper's \(n_{C_{4}}\) means the number of combinations \({}^{n}C_{4} = \frac{n(n-1)(n-2)(n-3)}{24}\). Setting this to 1365:

\[ \begin{aligned} n(n-1)(n-2)(n-3) &= 1365 \times 24 = 32760 \\ 15 \times 14 \times 13 \times 12 &= 32760 \end{aligned} \]

So \(n = 15\). Nearby values do not fit: \({}^{14}C_{4} = 1001\) and \({}^{16}C_{4} = 1820\).

Answer: A

The polygon has 20 sides.

Each of the \(n\) vertices joins to \(n - 3\) non-adjacent vertices, and each diagonal is counted twice:

\[ \begin{aligned} \frac{n(n - 3)}{2} &= 170 \\ n^{2} - 3n - 340 &= 0 \\ (n - 20)(n + 17) &= 0 \end{aligned} \]

So \(n = 20\) (the negative root is impossible). Check: \(\frac{20 \times 17}{2} = 170\). A 25-gon would have 275 diagonals and a 24-gon 252.