# Theorems

**Theorem**

^{n}P_{r}=^{n}C_{r }r!, 0 â‰¤ r â‰¤nThe number of combinations of n distinct things taken, 0 â‰¤ r â‰¤ n at a time is

**Proof**

Let the number of combinations of n things taken r at a time be denoted by x. Take one of these combinations. It contains r things which can be arranged among themselves in r! ways. Therefore, one combination gives rise to r! permutations. Therefore x combinations will give rise to x(r!) permutation. Also the number of permutations of n things taken r at a time is

^{ n}p

_{r}.

Thus

**Therefore,**

^{n}P_{r }=^{n}C_{r}r!, 0 â‰¤ r â‰¤n**Remarks:**

- The number of ways of selecting no object out of n distinct objects is clearly one, as we do not have to do anything in this case. Therefore
^{ n}C_{0 }= 1. This could have been derived directly from the formula for^{n}C_{r}. We have

^{n}C_{0 }= - The number of ways of selecting all the objects from the n distinct objects is 1, since there is only one way to choose all the objects from the n distinct objects. From the formula, we have

^{n}C_{n }= - The number of ways of selecting r objects out of n distinct objects is equal to the number of ways of selecting (n-r) objects out of n, that is
^{ n}C_{r }=^{n}C_{n-r}

**Proof**

Every time we take selection of r objects out of n distinct objects, we are left with a corresponding group of (n-r) objects. Therefore, the number of combinations of n objects taken r at a time is the same as the number of combinations of n objects taken (n-r) at a time.

^{n}C

_{r }=,0 â‰¤ r

**â‰¤**n

^{n}C

_{r }===

^{ n}C

_{r, }0 â‰¤ r

**â‰¤**n

Hence

^{ n}C

_{r }=

^{n}C

_{n-r}

- If
^{n}C_{a }=^{n}C_{b }then a = b or a = n - b

i.e., n = a + b

**Theorem**

^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}^{n}C

_{r}+

^{n}C

_{r-1}

=+

=+

=+

=+

=

^{ =n+1}C

_{r}