Question

15.

Answer 1

Answer:

a. selecting 2 from 3 using combination formula= 3way

b. selecting 1 from 3 using combination= 3way

Step-by-step explanation:

using combination formula that is:

3C2 = 3?÷(3-2)?2?= 3ways

3C1 = 3?÷(3-1)?1? = 3ways

b. n?= nx(n-1)(n-2)(n-3)(n-4)-------

           (

Answer 2

Explanation:

a)

${3 \choose 2} = \frac{3!}{2!(3-2)!} = \frac{6}{2*1} = 3$

${3 \choose 1} = \frac{3!}{1!(3-1)!} = \frac{6}{1*2} = 3$

b) First, we have that

${n \choose k} = \frac{n!}{k!(n-k)!}$

On the other hand,

${n \choose n-k} = \frac{n!}{(n-k)!(n(n-k))!} = \frac{n!}{(n-k)!k!}$

Therefore, both expressions are equal. This makes sense, because selecting k elements from a group of n is the same than specify which elements you will not select. In order to specify those elements you need to select the n-k elements that will not be selected. Hence, each time you are selecting k from n, you are also selecting n-k from n, and from that reason both combinatorial numbers are equal.