Question

Suppose we want to choose 3 letters, without replacement, from the 4 letters A, B, C, and D.

(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:

4 ways...

3 ways.......

Answer 2

Answer:

(a) number of ways of choosing 3 letters from the 4 letters, when the order of the choices is relevant is 24

(b)  number of ways of choosing 3 letters from the 4 letters, when the order of the choices is not relevant is 4

Step-by-step explanation:

given, there are  4 letters A, B, C, and D

(a) to choose 3 letters from the given 4, without replacement

also, the order of choice is important

we use permutations in this case

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

$4P_{3}$ = $\frac{4!}{(4-3)!}$

      = $\frac{4(3)(2)(1)}{1}$ = 24

number of ways of choosing 3 letters from the 4 letters, when the order of the choices is relevant is 24

(b) to choose 3 letters from the given 4, without replacement

also, the order of choice is not important

we use combination in this case

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

$4C_{3} = \frac{4!}{3!(4-3)!}$

        = $\frac{(4)3!}{3!}$ = 4

number of ways of choosing 3 letters from the 4 letters, when the order of the choices is not relevant is 4