Permutations & Combinations Questions and Answers

Answer: A

There are 461 such numbers.

Write every number with at most five digits as five digits with leading zeros. We need the number of solutions of \(d_{1} + d_{2} + \dots + d_{5} \leq 6\) with each \(d_{i} \geq 0\) (no digit can exceed 6 here, so the limit of 9 never binds). Adding a slack variable makes this an equation in six unknowns summing to 6:

\[ \binom{6 + 5}{5} = \binom{11}{5} = 462 \]

This count includes 00000, which is not a natural number, so the answer is \(462 - 1 = 461\). Option 462 is the trap of forgetting to remove zero.

Answer: A

There are 20 ways.

Choosing one envelope (5 ways) and one stamp (4 ways) are independent choices, so by the multiplication principle the number of ways to buy one of each is

\[ 5 \times 4 = 20 \]

9 is 5 + 4, the count of ways to buy an envelope or a stamp, not both, which is the common slip. 45 and 54 fit no step.

Answer: B

4 people can sit round a circular table in 3! ways.

In a circular arrangement, rotating everyone together gives the same seating, so one person's place is fixed and the rest are arranged relative to them.

\[ (n - 1)! = 3! = 6 \]

4! = 24 counts arrangements in a row, treating rotations as different. 4 and 1 undercount the arrangements.

For n people round a table where clockwise and anticlockwise orders count as the same, such as beads on a necklace, the count would be (n - 1)!/2.

Answer: A

ENCYCLOPAEDIA has thirteen letters: vowels E, O, A, E, A, I (six, with E and A each twice) and consonants N, C, Y, C, L, P, D (seven, with C twice).

Positions 2, 4, 6, 8, 10 and 12 are the six even ones - exactly enough for the vowels.

\[ \text{vowels} = \frac{6!}{2!\,2!} = \frac{720}{4} = 180 \]

\[ \text{consonants in the 7 odd positions} = \frac{7!}{2!} = 2520 \]

\[ 180 \times 2520 = 453{,}600 \]

The counts of vowels and even positions match exactly, which is what makes the arrangement possible at all.

Answer: C

Each question has five choices, answered independently.

\[ 5^{4} = 625 \text{ ways to answer the paper} \]

Exactly one of those has all four right, so

\[ 625 - 1 = 624 \]

The complement is the whole trick: counting the ways to get at least one wrong directly would need four separate cases.

Answer: C

Count all sequences and remove the forbidden ones.

\[ \text{total} = 7! = 5040 \]

\[ \text{youngest last} = 6! = 720 \]

\[ 5040 - 720 = 4320 \]

Equivalently, the last place has 6 acceptable players and the rest arrange freely: \( 6 \times 6! = 4320 \).

Answer: B

The two selections are independent, so their counts multiply.

\[ {}^{5}C_{2} \times {}^{6}C_{3} = 10 \times 20 = 200 \]

Order does not matter within either group, so combinations are used rather than permutations.

Answer: B

The claim is false. Circular permutations of \( n \) distinct objects number

\[ (n-1)! \]

not \( n! \). In a circle there is no fixed starting point, so the \( n \) rotations of any arrangement all look the same - fixing one object's position removes that redundancy and leaves \( (n-1)! \).

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: C

Count all pairs, then remove those that break the rule.

\[ \text{total} = {}^{5}C_{2} = 10 \]

Invalid pairs are those containing Mathematics 2 without Mathematics 1 - that is, Mathematics 2 paired with commerce, economics or statistics:

\[ 3 \text{ invalid pairs} \]

\[ 10 - 3 = 7 \]

The pair {Mathematics 1, Mathematics 2} is allowed, since the prerequisite is satisfied within the pair itself.