How do you evaluate 11C7?

Sep 22, 2017

See below.

Explanation:

The general formula for combinations is:

(n!)/(r!( n - r)!

Where $n$ is the number of objects and $r$ is how many are taken at a time.

n! means n x (n - 1) x ( n - 2 ).........( n - n + 2 )(n - n + 1):

This notation can seem confusing at first, but it basically just means the following:

5! => 5 xx 4 xx 3 xx 2 xx 1

7! =>7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1

etc.

Using formula:

$n = 11$
$r = 7$

(11!)/((7!(4)!

We can make the calculation easier by cancelling first:

$\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{\left(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\right) \left(4 \times 3 \times 2 \times 1\right)}$

$\frac{11 \times 10 \times 9 \times 8 \times \cancel{7} \times \cancel{6} \times \cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\left(\cancel{7} \times \cancel{6} \times \cancel{5} \times 4 \times 3 \times 2 \times 1\right) \left(\cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}\right)}$

This leaves us:

$\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \frac{7920}{24} = \textcolor{b l u e}{330}$

As a quick tip. Notice that what we are removing from the numerator is 7!

What remains in the denominator is 11 - 7 = 4.

This is the 4!#

We know this at the start because we had the 11 and the 7'

These are the quick ways of evaluating by hand. A calculator is much quicker, but where's the fun in that :)

Hope this helps you.