Question

suppose we want to choose 2 letters, without replacement, from the 5 letters A, B, C, D, and E (a) How many ways can this be done, if the order of the choices is relevant? (b) How many ways can this be done, if the order of the choices is not relevant?

Answer 1

Answer: a. 20  and b.10

Step-by-step explanation:

Given Letters = A, B, C, D, and E

Number of letters given = 5

a. When we choose 2 letters, without replacement from the 5 letters and order of the choices is relevant we use permutations.

Then, the number of ways we can choose 2 letters from 5 = $^5P_2$

$=\dfrac{5!}{(5-2)!}=\dfrac{5\times4\times3!}{3!}=20$  [∵ $^nP_r=\dfrac{n!}{(n-r)!}$]

So , when order of the choices is relevant , the total number of way to do this = 20

b. When we choose 2 letters, without replacement from the 5 letters and order of the choices is not relevant we use combinations.

Then, the number of ways we can choose 2 letters from 5 = $^5C_2$

$=\dfrac{5!}{2!(5-2)!}=\dfrac{5\times4\times3!}{2\times3!}=10$  [∵ $^nC_r=\dfrac{n!}{r!(n-r)!}$]

So , when order of the choices is not relevant , the total number of way to do this = 10