**Permutation and Combination practice Set 1**Advertisement

Here are few Permutation and Combination practice Set for you to prepare for your competitive exams.

### Example 1 – Permutation and Combination practice Set

In how many different way the word “HOPE” can be arranged so as vowel always come together ?

**Solution :** n = 4, r = 4 By formula : ^{4}P_{r} = 4! / (4 – 4)! = 4! / 0! = 4 x 3 x 2 x 1 / 1 = 24.

### Example 2 – Permutation and Combination practice Set

In how many different way the word “MATHEMATICS” can be arranged so as vowel always come together ?

**Solution :** In “MATHEMATICS” we treat the vowel AEAI as one letter So, without vowel 7 letter.

We have to arrange 7(MTHMTCS)letter + 1(AEAI) = 8 letter. and M occures twice and T occers twice rest are different.

Number of ways of arranging these letters = 8! / (2!) x (2!) = 10080.

Now, AEAI has 4 letters in which A occures 2 times and rest are different.

Number of ways of arrenging these letter = 4! / 2! = 12

Required number of words = 10080 x 12 = 120960.

### Example 3

In how many different way the word “COMPREHENSION” can be arranged so as vowel always come together ?

**Solution :** In “COMPREHENSION” we treat the vowel (OEEIO) as one letter So, without vowel 8 letter.

We have to arrange 8(CMPRHNSN)letter + 1(OEEIO) = 9 letter

8 Letter of which N occer 2 times and rst are different

Number of ways of arranging these letters = 9! / 2! = 181440 ways

Now vowel in which O and E occures 2 times can be arrange in 5! / 2! x 2! = 30 ways

Required number of ways =(181440 x 30) = 5443200.

### Example 4

In how many different way the word “ENGINEERING” can be arranged so as vowel always come together ?

**Solution :** In “ENGINEERING” contains 11 letters, where 3E 3N, 2G, 2I and 1R.

So Required number of arrangements = 11! / (3!)(3!)(2!)(2!)(1!) = 11x10x9x8x7x6x5x4x3x2x1 / 3x1x3x1x2x1x2x1x1x1 = 277200.

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I have a burning question here and would appreciate if you can help:

In how many ways can a group of 3 men be selected from 7 men? How many ways of selection are there if one of two particular men must not be included?