Question

Find the value of each of these quantities:

a) C (5,1)

b) C (5,3)

c) C (8,4)

d) C (8,8)

e) C (8,0)

f) C (12,6)

Short answer

(a) The resultant answer is 5.

(b) The resultant answer is 10.

(c) The resultant answer is 70.

(d) The resultant answer is 1.

(e) The resultant answer is 1.

(f) The resultant answer is 924.

Step-by-step solution

Given data

The given data is the quantities.

Concept of Permutation

Definition permutation (order is important):$\mathbf{P}\left(n,r\right)\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}}$

with n! = n . (n - 1). ..... 2 . 1 .

Evaluate the definition of a combination

(a)

We need to use the definition of permutation.

n = 5

r = 1

Evaluate the definition of a permutation:

$\begin{array}{r}C(5,1)=\frac{5!}{1!(5-1)!}\\ =\frac{5!}{1!4!}\\ =5\end{array}$

Evaluate the definition of a combination

(b)

We need to use the definition of permutation.

n = 5

r = 3

Evaluate the definition of a permutation:

$\begin{array}{r}C(5,3)=\frac{5!}{3!(5-3)!}\\ =\frac{5!}{3!2!}\\ =10\end{array}$

Evaluate the definition of a combination

(c)

We need to use the definition of permutation.

n = 8

r = 4

Evaluate the definition of a permutation:

$\begin{array}{r}C(8,4)=\frac{8!}{4!(8-4)!}\\ =\frac{8!}{4!4!}\\ =70\end{array}$

Evaluate the definition of a combination

(d)

We need to use the definition of permutation.

n = 8

r = 8

Evaluate the definition of a permutation:

$\begin{array}{r}C(8,8)=\frac{8!}{8!(8-8)!}\\ =\frac{8!}{8!0!}\\ =1\end{array}$

Evaluate the definition of a combination

(e)

We need to use the definition of permutation.

n = 8

r = 0

Evaluate the definition of a permutation:

$\begin{array}{r}C(8,0)=\frac{8!}{0!(8-0)!}\\ =\frac{8!}{0!8!}\\ =1\end{array}$

Evaluate the definition of a combination

(f)

We need to use the definition of permutation.

n = 12

r = 6

Evaluate the definition of a permutation:

$\begin{array}{r}C(12,6)=\frac{12!}{6!(12-6)!}\\ =\frac{12!}{6!6!}\\ =924\end{array}$