Question

If three alphabets are to be chosen from a, b, c, d and e such that repetition is not allowed then in how many ways it can be done?

Answer 1

We can choose 3 different letters from 5 in ⁵C₃ ways.

Also we can arrange these 3 letters in 3! ways.

So ways to choose = 10

Ways to arrange = 10 * 3! = 60

Answer 2

There are total 60 possible ways to choose any 3 alphabets from the given alphabets without repetition.

Explanation:

• We have total 5 alphabets.

• For choosing 3 alphabets out of these 5, we have to apply permutations on it.

Number of permutations =  $5P3$

$nPr =\frac{ n!}{r!(n-r)!}$

$5P3 = \frac{5!}{3!(5-3)!}$

   $= \frac{5!}{3!2!}$

   $=\frac{4\times 5}{2}$

   =10

• Now we can arrange all these 3 alphabets in different orders.

• So, number of arrangements = 3! = 6

• So, Total number of different selections = 6*10

                                                                           = 60

Hence we can choose 3 alphabets in 60 different ways from given 5 alphabets.