2.4 MCQs-Permutation and Combination

Permutation and Combination MCQs

Fundamental Counting Principle

1. If there are 3 ways to travel from City A to City B, and 4 ways to travel from City B to City C, how many different ways are there to travel from City A to City C via City B?

1. 7

2. 12

3. 4

4. 3

Show me the answer

Answer: 2. 12

Explanation:

• This uses the Fundamental Counting Principle (Multiplication Principle).

• If one event can occur in $m$ ways and a second independent event can occur in $n$ ways, then the two events together can occur in $m \times n$ ways.

• Here: Ways from A to B = 3, Ways from B to C = 4.

• Total ways from A to C via B = $3 \times 4 = 12$.

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Permutations

2. A permutation is an arrangement of objects where:

1. Order does not matter

2. Order matters

3. Repetition is always allowed

4. Objects are identical

Show me the answer

Answer: 2. Order matters

Explanation:

• The key difference between permutations and combinations is order.

• Permutations consider different orders/arrangements as different outcomes.

• Combinations consider different orders as the same outcome.

• Example: The arrangements AB and BA are different permutations but the same combination.

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3. The number of permutations of n distinct objects taken r at a time is given by:

1. $\frac{n!}{r!}$

2. $\frac{n!}{(n-r)!}$

3. $\frac{n!}{r!(n-r)!}$

4. $n^r$

Show me the answer

Answer: 2. $\frac{n!}{(n-r)!}$

Explanation:

• The formula for permutations (without repetition) is: $P(n, r) = \frac{n!}{(n-r)!}$ where $n! = n \times (n-1) \times ... \times 2 \times 1$.

• This counts the number of ways to arrange r objects out of n distinct objects.

• $P(n, n) = n!$ (arranging all n objects).

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4. How many different 4-letter codes can be formed from the letters A, B, C, D, E if no letter is repeated?

1. 120

2. 24

3. 625

4. 20

Show me the answer

Answer: 1. 120

Explanation:

• This is a permutation problem: We are arranging 4 letters out of 5 distinct letters, order matters, no repetition.

• Use $P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = \frac{120}{1} = 120$.

• Alternatively: For the first position: 5 choices, second: 4 choices, third: 3 choices, fourth: 2 choices. Total = $5 \times 4 \times 3 \times 2 = 120$.

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5. The number of ways to arrange the letters in the word "MATH" is:

1. 4

2. 16

3. 24

4. 256

Show me the answer

Answer: 3. 24

Explanation:

• "MATH" has 4 distinct letters.

• The number of permutations of n distinct objects is $n!$.

• Here, $n = 4$, so number of arrangements = $4! = 4 \times 3 \times 2 \times 1 = 24$.

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Combinations

6. A combination is a selection of objects where:

1. Order does not matter

2. Order matters

3. Repetition is always allowed

4. Objects must be distinct

Show me the answer

Answer: 1. Order does not matter

Explanation:

• Combinations are used when we are selecting items and the order of selection is NOT important.

• Example: Choosing a committee of 3 people from a group of 10. The committee {Alice, Bob, Charlie} is the same as {Bob, Charlie, Alice}.

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7. The number of combinations of n distinct objects taken r at a time is given by:

1. $\frac{n!}{r!}$

2. $\frac{n!}{(n-r)!}$

3. $\frac{n!}{r!(n-r)!}$

4. $n^r$

Show me the answer

Answer: 3. $\frac{n!}{r!(n-r)!}$

Explanation:

• The formula for combinations (without repetition) is: $C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

• This is also called "n choose r".

• It counts the number of ways to choose r objects from n, disregarding order.

• Relationship with permutations: $P(n, r) = C(n, r) \times r!$ because you choose (combination) then arrange (r! ways).

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8. How many different committees of 3 people can be formed from a group of 7 people?

1. 210

2. 35

3. 343

4. 21

Show me the answer

Answer: 2. 35

Explanation:

• This is a combination problem because the order in which committee members are chosen does not matter.

• Use $C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!}$.

• Calculate: $\frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35$.

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9. The value of $\binom{5}{2}$ is:

1. 10

2. 20

3. 25

4. 120

Show me the answer

Answer: 1. 10

Explanation:

• $\binom{5}{2} = C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!}$.

• Calculate: $\frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$.

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Permutations with Repetition

10. The number of different permutations of the letters in the word "MISSISSIPPI" is:

1. $11!$

2. $\frac{11!}{4!4!2!}$

3. $\frac{11!}{4!4!}$

4. $\frac{11!}{2!}$

Show me the answer

Answer: 2. $\frac{11!}{4!4!2!}$

Explanation:

• When arranging letters where some are repeated, we use the formula: $\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$ where $n$ is total letters, and $n_1, n_2, ...$ are counts of each repeating letter.

• For "MISSISSIPPI": n = 11 letters. M: 1, I: 4, S: 4, P: 2.

• Number of distinct arrangements = $\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{11!}{4!4!2!}$.

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11. How many distinct 3-digit numbers can be formed using the digits 1, 2, 3, 4 if repetition of digits is allowed?

1. 24

2. 64

3. 12

4. 81

Show me the answer

Answer: 2. 64

Explanation:

• Since repetition is allowed and we are forming a sequence (order matters), we use the multiplication principle.

• For a 3-digit number: Hundreds place: 4 choices, Tens place: 4 choices, Ones place: 4 choices.

• Total numbers = $4 \times 4 \times 4 = 4^3 = 64$.

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Combinations with Repetition

12. The formula for the number of combinations of n objects taken r at a time, with repetition allowed, is:

1. $\binom{n}{r}$

2. $\binom{n+r-1}{r}$

3. $\frac{n!}{r!(n-r)!}$

4. $\frac{(n+r-1)!}{r!}$

Show me the answer

Answer: 2. $\binom{n+r-1}{r}$

Explanation:

• This is often called "stars and bars" or combinations with replacement.

• The formula is: $\binom{n+r-1}{r} = \frac{(n+r-1)!}{r!(n-1)!}$.

• Example: Choosing 3 scoops of ice cream from 5 flavors, where you can choose the same flavor more than once.

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Circular Permutations

13. The number of ways to arrange n distinct objects in a circle is:

1. $n!$

2. $(n-1)!$

3. $\frac{n!}{2}$

4. $(n-1)!/2$

Show me the answer

Answer: 2. $(n-1)!$

Explanation:

• In circular permutations, rotations are considered the same arrangement.

• To arrange n distinct objects in a circle, fix one object's position to account for rotational symmetry.

• Then arrange the remaining (n-1) objects in the remaining spots: $(n-1)!$ ways.

• Example: 5 people around a round table can be arranged in $(5-1)! = 4! = 24$ ways.

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Relationship and Properties

14. Which of the following is always true?

1. $P(n, r) = C(n, r)$

2. $C(n, r) = C(n, n-r)$

3. $P(n, r) = P(n, n-r)$

4. $C(n, 0) = 0$

Show me the answer

Answer: 2. $C(n, r) = C(n, n-r)$

Explanation:

• This is the symmetry property of combinations.

• Choosing r objects from n is the same as choosing the (n-r) objects to leave behind.

• $C(n, r) = \frac{n!}{r!(n-r)!}$ and $C(n, n-r) = \frac{n!}{(n-r)!r!}$, so they are equal.

• Example: $C(5, 2) = C(5, 3) = 10$.

• Note: $C(n, 0) = 1$ (one way to choose nothing).

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15. If $C(n, 5) = C(n, 8)$, then what is the value of n?

1. 5

2. 8

3. 13

4. 3

Show me the answer

Answer: 3. 13

Explanation:

• Using the symmetry property $C(n, r) = C(n, n-r)$.

• If $C(n, 5) = C(n, 8)$, then either:

  1. $5 = 8$ (impossible), or

  2. $5 = n - 8$, which gives $n = 5 + 8 = 13$.

• Check: $C(13, 5) = C(13, 8)$ because 13 - 5 = 8.

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Application Problems

16. In how many ways can a committee of 4 men and 3 women be chosen from a group of 7 men and 5 women?

1. $C(7,4) + C(5,3)$

2. $C(7,4) \times C(5,3)$

3. $C(12,7)$

4. $P(7,4) \times P(5,3)$

Show me the answer

Answer: 2. $C(7,4) \times C(5,3)$

Explanation:

• This involves multiple independent choices. Use the multiplication principle.

• Choose 4 men from 7: $C(7, 4)$ ways.

• Choose 3 women from 5: $C(5, 3)$ ways.

• These choices are independent, so multiply: Total ways = $C(7,4) \times C(5,3)$.

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17. How many different 5-card hands can be dealt from a standard deck of 52 cards?

1. $P(52, 5)$

2. $52^5$

3. $C(52, 5)$

4. $\frac{52!}{5!}$

Show me the answer

Answer: 3. $C(52, 5)$

Explanation:

• In a card hand, the order in which cards are received does not matter.

• Therefore, it's a combination problem.

• Number of 5-card hands = $C(52, 5) = \frac{52!}{5! \cdot 47!}$.

• This is a very large number, approximately 2.6 million.

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18. A password consists of 2 letters followed by 3 digits. How many different passwords are possible if letters and digits can be repeated?

1. $26^2 \times 10^3$

2. $P(26,2) \times P(10,3)$

3. $C(26,2) \times C(10,3)$

4. $26 \times 25 \times 10 \times 9 \times 8$

Show me the answer

Answer: 1. $26^2 \times 10^3$

Explanation:

• Since repetition is allowed, use the multiplication principle for each position.

• For the 2 letters: 26 choices for first letter, 26 for second = $26^2$.

• For the 3 digits: 10 choices for each digit = $10^3$.

• Total passwords = $26^2 \times 10^3 = 676 \times 1000 = 676,000$.

• If repetition were NOT allowed, it would be $P(26,2) \times P(10,3)$.

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Distinguishable vs. Identical Objects

19. In how many ways can 10 identical chocolates be distributed among 4 children?

1. $C(10,4)$

2. $P(10,4)$

3. $C(13,3)$

4. $4^{10}$

Show me the answer

Answer: 3. $C(13,3)$

Explanation:

• This is a classic "stars and bars" / combinations with repetition problem.

• We are distributing n identical items (10 chocolates) into k distinct groups (4 children).

• The formula is: $C(n + k - 1, k - 1) = C(10 + 4 - 1, 4 - 1) = C(13, 3)$.

• This counts the number of non-negative integer solutions to $x_1 + x_2 + x_3 + x_4 = 10$, where each $x_i$ is the number of chocolates a child gets.

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20. The number of ways to choose a president, vice-president, and secretary from a club of 10 members is:

1. $C(10,3)$

2. $P(10,3)$

3. $10^3$

4. $3^{10}$

Show me the answer

Answer: 2. $P(10,3)$

Explanation:

• Here, order matters because the positions (president, vice-president, secretary) are distinct.

• Choosing Alice as president and Bob as vice-president is different from choosing Bob as president and Alice as vice-president.

• Therefore, this is a permutation problem: arranging 3 people out of 10 into specific roles.

• Number of ways = $P(10,3) = \frac{10!}{7!} = 10 \times 9 \times 8 = 720$.