1000 + Permutations and Combinations Aptitude Questions

This section will cover Multiple Choice Questions on Permutations and Combinations under Quantitative aptitude or Numerical ability. Target should be to complete all Questions with 80% accuracy. Most of the questions covered in this section has appeared in recent competitive exams. Regular practice of these sample questions should help you in achieving a good Test score.

Basic Concepts of Permutations and Combinations

We have covered few tips for solving Aptitude Questions on Permutations and Combinations. Students should go through them before attempting Test series.

• Permutations formula: Permutations is defined as arrangement of r things that can be done out of total n things. This is denoted by ⁿP_r which is equal to n!/(n-r)!
• Combinations formula
  1. Combinations is defined as selection of r things that can be done out of total n things. This is denoted by ⁿC_r which is equal to n!/r!(n-r)!
  2. As per the Fundamental Principle of Counting, if a particular thing can be done in m ways and another thing can be done in n ways, then either one of the two can be done in m + n ways and both of them can be done in m × n ways.

Solved Permutations and Combinations Problems

Example 1: How many four-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 (Repetition of digits not allowed)?

Solution: Thousand’s place can be filled in 6 ways. Hundred’s place can be filled in 5 ways. Ten’s place can be filled in 4 ways. Unit’s place can be filled in 3 ways. So, using the Fundamental Principle of Counting, we get the answer as 6 × 5 × 4 × 3 = 360. Or using the formula of Permutations, we need to arrange 4 digits out of total 6 digits. This can be done in ⁶P₄= 360 ways.

Example 2: A person has 6 friends to be invited for dinner through invitation cards, and he has 3 servants. In how many ways can he extend the invitation card?

Solution: We can see that the 1^st friend has 3 options to receive the card, i.e. either from 1^st servant or 2^nd or 3^rd. Similarly, 2nd friend also has 3 options to receive the card, i.e. either from 1^st servant or 2^nd or 3^rd. So we can say that each of the 6 friends has 3 options to receive the card. Hence the answer would be 3 × 3 × 3 × 3 × 3 × 3 = 3⁶=729 ways.

Example 3: How many words can be formed from the letters of the word TRIANGLE with T always at the beginning and E at the end.

Solution: If we fix up T in the beginning and E at the end, then the remaining 6 letters can be arranged in ⁶P₆ that is 6! which is 720 ways.

Example 4: How many words can be formed with the letters of the word “ORDINATE” so that vowels occupy odd places ?

Solution : There are 4 vowels and 4 consonants in the word “ORDINATE”. We have to arrange 8 letters in such a way that vowels occupy odd places i.e. 1,3,5 and 7. So four vowels can be arranged in these 4 odd places as ⁴P₄ or 4! . And remaining 4 consonants can occupy 4 positions left in ⁴P₄ way.

So, total ways or permutations will be 4! X 4! = 576

Example 5: In how many ways 5 boys and 3 girls can be seated in a row that no 2 girls are together ?

Solution: 5 boys can be seated in ⁵P₅ ways. So now 3 girls can seat in between boys so that they are not together.

__ B __ B __ B __ B __ B__

So now 3 girls can occupy 6 places which is ⁶P₃ ways.

Hence total ways are ⁵P₅ * ⁶P₃ = 14400

Example 6: How many 3 digit numbers between 400 and 1000 can be formed with the digits 0, 2, 3, 4, 5, 6 if no digit is repeated ?

Solution: Since the number is greater than 400, so hundred’s place can be filled by 4 , 5 or 6 only (3 ways) . Now unit and ten’s place can be filled by remaining 5 digits i.e. ⁵P₂.

so total ways will be 5 * ⁵P₂ = 60

Example 7: How many different words can be formed with the letters of the word MISSISSIPPI ?

Solution: Total 11 characters are there in the word. Out of which 4 S, 4 I, 2 P are repeated. So in such problems we will be dividing total ways by repeating ways i.e 11!/(4!)(4!)(2!) = 34650

Example 8: In how many ways a committee of 5 members can be selected from 6 men and 5 women, consisting of 3 men and 2 women ?

Solution 8: 3 men out of 6 can be selected in ⁶C₃ ways. 2 women out of 5 ways can be selected in ⁵C₂ ways. So ⁶C₃ * ⁵C₂ = 200 ways.

Example 9: A team of 3 members need to be formed out of 5 men and 2 women. In how many ways team can be formed so as to include at least 1 woman ?

Solution 9: Team of 3 can be formed in 2 ways

a) 2 men and 1 woman – ⁵C₂ * ²C₁

b) 1 man and 2 women – ⁵C₁ * ²C₂

So total ways will be (⁵C₂ * ²C₁) + (⁵C₁ * ²C₂) = 25 ways.

Permutations and Combinations MCQ –

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For more Numerical ability or Quantitative Aptitude Questions, please refer here.