Question

Evaluate C(7, 7). a) 0 b) 1 c) 7

Answer 1

We have to find the value of C(7,7)

We will evaluate this using the combination formula which states:

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

Putting the value n = r= 7

we get,

$C(7,7)=^7C_{7}=\frac{7!}{7!(7-7)!}$

= $\frac{7!}{7!(0)!}$

Since the value of 0! is 1

So, C(7,7)= $\frac{7!}{7!}$

C(7,7)=1

Therefore, the value of C(7,7) is 1.

So, Option B is the correct answer.

Answer 2

By direct evaluation, we will see that C(7, 7) = 1, so the correct option is b.

How to evaluate C(N, K)?

When we have N elements, the number of different sets of K elements (such that K ≤ N) is given by:

$C(N, K) = \frac{N!}{(N - K)!*K!}$

Here we just need to evaluate it in N = 7 and K = 7, using logic, if we have a set of 7 elements (N = 7) we could only make one set of 7 elements out of it (K = 7), but let's do the evaluation.

$C(7, 7) = \frac{7!}{(7 - 7)!7!} = \frac{7!}{7!} = 1$

Where we used that 0! = 1.

So the correct option is b, C(7, 7) = 1.

If you want to learn more about combinations, you can read:

https://brainly.com/question/251701