Question

Prove the following relation:
$^{n}{{C}_{r}}={{\text{ }}^{n}}{{C}_{n-r}}$

Answer 1

Hint: In order to prove this expression, we should know that the combination can be defined as the number of ways of combining things irrespective of their orders. And it can be expressed as $^{n}{{C}_{r}}$ in mathematical terms where n represents the total number of choices and r represents the number of choices to be combined. The mathematical formula of combining r choices from n choices is:
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Complete step-by-step answer:
In this question, we have to prove that
$^{n}{{C}_{r}}={{\text{ }}^{n}}{{C}_{n-r}}$
To prove this expression, we will first consider LHS, that is,
$LHS={{\text{ }}^{n}}{{C}_{r}}$
Now, we know that $^{n}{{C}_{r}}$ can be explained as the number of ways of combining r choices from the total number of n choices irrespective of their orders and can be expanded by using the formula
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Therefore, we will get LHS as
$LHS=\dfrac{n!}{r!\left( n-r \right)!}......\left( i \right)$
Now, we will consider RHS of the expression, that is
$RHS={{\text{ }}^{n}}{{C}_{n-r}}$
Now, we will again use the formula that is
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
And here, we will put r = n – r. So, we can write RHS as,
$RHS=\dfrac{n!}{\left( n-r \right)!\left( n-\left( n-r \right) \right)!}$
$RHS=\dfrac{n!}{\left( n-r \right)!\left( n-n+r \right)!}$
$RHS=\dfrac{n!}{\left( n-r \right)!r!}....\left( ii \right)$
Now, we can see that from equation (i) and (ii),
LHS = RHS
$^{n}{{C}_{r}}={{\text{ }}^{n}}{{C}_{n-r}}$
Hence proved

Note: We can also do this question by putting r = n – r in the equation (i) and then we can write that as $\dfrac{n!}{\left( n-r \right)!\left( n-n+r \right)!}$ which can be further written as $\dfrac{n!}{\left( n-r \right)!\left( n-\left( n-r \right) \right)!}$ which is equal to $^{n}{{C}_{n-r}}$.