If for 3 r 30, (^30C_30-r)+3(^30C_31-r)+3(^30C_32-r)+(^30C_33-r)=^mC_r, then m equals :

Question

If for $3\le r\le 30$, $({}^{30}{C}_{30-r})+3({}^{30}{C}_{31-r})+3({}^{30}{C}_{32-r})+({}^{30}{C}_{33-r})={}^{m}{C}_{r}$, then m equals :

Accepted Answer

Given

${}^{30}{C}_{30-r}+3{}^{30}{C}_{31-r}+3{}^{30}{C}_{32-r}+{}^{30}{C}_{33-r}={}^{m}{C}_{r}$

for $3\le r\le 30$.

We have to find $m$.

First, use the symmetry property of combinations:

${}^{n}{C}_{k}={}^{n}{C}_{n-k}$

So,

${}^{30}{C}_{30-r}={}^{30}{C}_{r}$

${}^{30}{C}_{31-r}={}^{30}{C}_{r-1}$

because

$30-(31-r)=r-1$

Similarly,

${}^{30}{C}_{32-r}={}^{30}{C}_{r-2}$

and

${}^{30}{C}_{33-r}={}^{30}{C}_{r-3}$

Therefore the left-hand side becomes

${}^{30}{C}_{r}+3{}^{30}{C}_{r-1}+3{}^{30}{C}_{r-2}+{}^{30}{C}_{r-3}$

Now notice the coefficients:

$1,3,3,1$

These are the binomial coefficients in the expansion of

$(1+x{)}^{3}$

Also, by Vandermonde's identity,

${}^{30}{C}_{r}+3{}^{30}{C}_{r-1}+3{}^{30}{C}_{r-2}+{}^{30}{C}_{r-3}={}^{33}{C}_{r}$

Let us see this through binomial expansion:

Coefficient of ${x}^{r}$ in $(1+x{)}^{33}$ is

${}^{33}{C}_{r}$

But

$(1+x{)}^{33}=(1+x{)}^{30}(1+x{)}^{3}$

and

$(1+x{)}^{3}=1+3x+3{x}^{2}+{x}^{3}$

So coefficient of ${x}^{r}$ in $(1+x{)}^{30}(1+3x+3{x}^{2}+{x}^{3})$ is

${}^{30}{C}_{r}+3{}^{30}{C}_{r-1}+3{}^{30}{C}_{r-2}+{}^{30}{C}_{r-3}$

Hence,

${}^{30}{C}_{r}+3{}^{30}{C}_{r-1}+3{}^{30}{C}_{r-2}+{}^{30}{C}_{r-3}={}^{33}{C}_{r}$

Comparing with

${}^{m}{C}_{r}$

we get

$m=33$

The correct answer is:

$33$

So, Option C.

Solutions · 1