How do you evaluate 11C7?

1 Answer
Sep 22, 2017

Answer:

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:

#( 11 xx 10 xx 9 xx 8 xx 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1)/(( 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1 )( 4 xx 3 xx 2 xx 1))#

#( 11 xx 10 xx 9 xx 8 xx cancel(7) xx cancel(6) xx cancel(5) xx cancel(4) xx cancel(3) xx cancel(2) xx cancel(1))/(( cancel(7) xx cancel(6) xx cancel(5) xx 4 xx 3 xx 2 xx 1 )( cancel(4) xx cancel(3) xx cancel(2) xx cancel(1)))#

This leaves us:

#(11 xx 10 xx 9 xx 8)/( 4 xx 3 xx 2 xx 1 ) = 7920/24 = color(blue)(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.