Home > Math Shortcuts > Permutation and Combination practice Set 1
Permutation and Combination practice Set 1
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 : 4Pr = 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.
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.
We provide few shortcut tricks on this topic. Please visit this page to get updates on more Math Shortcut Tricks. You can also like our facebook page to get updates.
If you have any question regarding this topic then please do comment on below section. You can also send us message on facebook.
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?