Question

How do you evaluate ${}^9{C_3}$?

Answer 1

Hint: A combination is a way to order or arrange a set or number of things . We have to use this formula for getting this value. The formula for a combination of choosing r unique ways from n possibilities are mentioned below.
Where n is the number of items and r is the unique arrangement.
Formula Used: ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete step by step answer:
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
${}^9{C_3} = \dfrac{{9!}}{{3!\left( {9 - 3} \right)!}}$
$ = \dfrac{{9!}}{{3!6!}}$
$ = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3!6!}}$
$ = \dfrac{{9 \times 8 \times 7 \times {6}!}}{{{3}!{6}!}}$
 $ = \dfrac{{{{{9}}^3} \times {{{8}}^4} \times 7}}{{{3} \times {2} \times 1}}$
 $ = 3 \times 4 \times 7 = 84$

Note:By just plugging the numbers in the given formula we can get the required value.
Remember from our factorial lesson that $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right)....... \times 2 \times 1$.
Here for solving these question we
first, calculate the numerator then,
calculate denominator then,
calculate the second denominator $r!$.
Finally calculate our combination value ${}^n{C_r}$
There are two types of combinations (remember the order doesn't matter now)
Combinations with Repetition
Actually, these are the hardest to explain, so we will come back to this later.
Combinations without Repetition
This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win!
The easiest way to explain it is to:
assume that the order does matter (ie permutations),
then alter it so the order does not matter.