In how many ways can the letters of the word CAT be arranged?

6. Three distinct letters can be arranged in 3! = 6 ways.

In how many ways can a committee of 3 be chosen from 8 people?

56. Order does not matter: ^8C3 = 8!3!\,5! = 56.

How many 3-digit numbers can be formed from the digits 1 to 9 if no digit is repeated?

504. The hundreds digit has 9 choices, the tens 8 and the units 7: 9 8 7 = 504.

In how many distinct ways can the letters of the word LEVEL be arranged?

30. LEVEL has 5 letters with L and E each repeated twice: 5!2!\,2! = 30.

From 7 men and 4 women, in how many ways can a committee of 5 be formed containing exactly 2 women?

210. Choose 2 women from 4 and 3 men from 7: ^4C2 ^7C3 = 6 35 = 210.

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Quantitative Aptitude · Counting, Probability, Calendar & Clocks

Permutations and Combinations — Questions & Answers

Use a permutation when order matters (arrangements, seat numbers, forming words) and a combination when it does not (committees, choosing a subset). The multiplication principle underlies both.

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Key formulas included

Key formulas

$^nP_r = \dfrac{n!}{(n-r)!}$ (arrangements); $\;^nC_r = \dfrac{n!}{r!\,(n-r)!}$ (selections)
$^nC_r = {}^nC_{n-r}$ and $\;^nC_0 + {}^nC_1 + \dots + {}^nC_n = 2^n$
Circular arrangements of $n$ distinct items: $(n-1)!$

5 solved Permutations and Combinations questions

Try each question first — tap an option to check your answer, or open Learn step by step for a guided walkthrough.

01
In how many ways can the letters of the word CAT be arranged?
5 steps · reveal one at a time
1. Step 1: Identify the task
  We need to find out all the possible unique ways to arrange the letters of the word CAT. This is a classic permutation problem.
2. Step 2: Count distinct letters
  First, let's look at the word CAT. It has three letters: C, A, and T. All these letters are distinct, meaning each one is different from the others.
3. Step 3: Apply permutation rule
  When you have $n$ distinct items and you want to arrange all of them, the number of possible arrangements is given by $n!$ (n factorial). Here, we have $n=3$ distinct letters.
4. Step 4: Calculate the arrangements
  To find the number of ways, we calculate $3!$ . This means $3 \times 2 \times 1$ . Performing this multiplication gives us $6$ .
5. Step 5: The answer
  So the answer is (C) 6. There are 6 distinct ways to arrange the letters C, A, and T, which are CAT, CTA, ACT, ATC, TCA, and TAC.
02
In how many ways can a committee of 3 be chosen from 8 people?
5 steps · reveal one at a time
1. Step 1: Identify the Goal
  We need to find the number of ways to choose a committee of 3 people from a group of 8. For a committee, the order in which people are selected does not matter.
2. Step 2: Use Combinations
  Since the order of selection doesn't matter, this is a combination problem. We use the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$ , where $n$ is the total number of items and $r$ is the number to choose.
3. Step 3: Calculate the Value
  Substitute $n=8$ and $r=3$ into the formula: $^8C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$ . This expands to $\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(5 \times 4 \times 3 \times 2 \times 1)}$ . We can simplify this.
4. Step 4: Simplify and Solve
  Cancel out the $5!$ terms: $^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$ . This simplifies to $\frac{336}{6}$ , which equals $56$ . So there are 56 ways to choose the committee.
5. Step 5: The answer
  So the answer is (B) 56, because there are 56 unique combinations of 3 people that can be chosen from a group of 8.
03
How many 3-digit numbers can be formed from the digits 1 to 9 if no digit is repeated?
5 steps · reveal one at a time
1. Step 1: Identify the task
  We need to form 3-digit numbers using digits from 1 to 9. A key condition is that no digit can be repeated. This means each position in the 3-digit number must use a unique digit.
2. Step 2: Choose the first digit
  For the hundreds digit, we have 9 available choices (digits 1 through 9). Since no digit can be repeated, once we pick a digit for this position, it cannot be used again.
3. Step 3: Choose remaining digits
  After choosing the hundreds digit, 8 digits remain. So, for the tens digit, we have 8 choices. Then, for the units digit, 7 digits remain, giving us 7 choices. Each choice reduces the available digits.
4. Step 4: Calculate total possibilities
  To find the total number of 3-digit numbers, we multiply the number of choices for each position: $9 \times 8 \times 7$ . This calculation gives us the total unique combinations possible.
5. Step 5: The answer
  So the answer is (D) 504. We calculate the total number of unique 3-digit numbers by multiplying the number of choices for each position: $9 \times 8 \times 7 = 504$ .
04
In how many distinct ways can the letters of the word LEVEL be arranged?
5 steps · reveal one at a time
1. Step 1: Analyze the word
  First, let's look at the word given: LEVEL. We need to count the total number of letters and identify any letters that are repeated. This will help us determine the correct method for counting arrangements.
2. Step 2: Count letters and repetitions
  The word LEVEL has a total of 5 letters. Let's count how many times each unique letter appears: 'L' appears 2 times, 'E' appears 2 times, and 'V' appears 1 time. These repetitions are key.
3. Step 3: Apply permutation formula
  To find the number of distinct arrangements for a set of items with repetitions, we use the formula for permutations with repetitions: $\frac{n!}{n_1!n_2!...n_k!}$ , where $n$ is the total number of items, and $n_1, n_2, ..., n_k$ are the counts of each repeated item.
4. Step 4: Calculate the arrangements
  Using our formula, $n=5$ (total letters), $n_L=2$ (L repeated twice), and $n_E=2$ (E repeated twice). So, the calculation is $\frac{5!}{2!2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30$ . There are 30 distinct ways.
5. Step 5: The answer
  So the answer is (B) 30. This is the total number of unique arrangements possible for the letters in the word LEVEL, accounting for the repeated letters 'L' and 'E'.
05
From 7 men and 4 women, in how many ways can a committee of 5 be formed containing exactly 2 women?
5 steps · reveal one at a time
1. Step 1: Understand the Goal
  We need to form a committee of 5 people from a group of 7 men and 4 women. The key condition is that the committee must contain exactly 2 women.
2. Step 2: Determine Committee Makeup
  If the committee has exactly 2 women, and the total committee size is 5, then the remaining $5 - 2 = 3$ members must be men. So, we need to choose 2 women and 3 men.
3. Step 3: Calculate Women's Choices
  First, find the number of ways to choose 2 women from the 4 available women. This is a combination, calculated as $^4C_2$ . $^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$ ways.
4. Step 4: Calculate Men's Choices
  Next, find the number of ways to choose 3 men from the 7 available men. This is also a combination, calculated as $^7C_3$ . $^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ ways.
5. Step 5: The answer
  To find the total number of ways to form the committee, multiply the ways to choose women by the ways to choose men: $6 \times 35 = 210$ . So the answer is (C) 210, because this is the total number of unique committees possible under the given conditions.

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