The number of elements in the set S=(r,k):k Z^36C_r+1=6(^35C_r)(k^2-3) is :

Question

The number of elements in the set $S={(r,k):k\in Z{}^{36}{C}_{r+1}=\frac{6({}^{35}{C}_{r})}{({k}^{2}-3)}}$ is :

Accepted Answer

We need to find the number of ordered pairs $(r,k)$ satisfying

${}^{36}{C}_{r+1}=\frac{6({}^{35}{C}_{r})}{{k}^{2}-3},$

where $k\in Z$.

Let us simplify the relation using the standard NCERT combination formula.

We know that

${}^{n}{C}_{r+1}=\frac{n!}{(r+1)!(n-r-1)!}$

and

${}^{n-1}{C}_{r}=\frac{(n-1)!}{r!(n-r-1)!}.$

So,

$\frac{{}^{36}{C}_{r+1}}{{}^{35}{C}_{r}}=\frac{\frac{36!}{(r+1)!(35-r)!}}{\frac{35!}{r!(35-r)!}}=\frac{36!}{35!}\cdot \frac{r!}{(r+1)!}=36\cdot \frac{1}{r+1}=\frac{36}{r+1}.$

Hence,

${}^{36}{C}_{r+1}={}^{35}{C}_{r}\cdot \frac{36}{r+1}.$

Given,

${}^{36}{C}_{r+1}=\frac{6({}^{35}{C}_{r})}{{k}^{2}-3}.$

Comparing both expressions,

${}^{35}{C}_{r}\cdot \frac{36}{r+1}=\frac{6({}^{35}{C}_{r})}{{k}^{2}-3}.$

Since ${}^{35}{C}_{r}
e 0$, cancel it:

$\frac{36}{r+1}=\frac{6}{{k}^{2}-3}.$

Cross-multiplying,

$36({k}^{2}-3)=6(r+1).$

Divide by $6$:

$6({k}^{2}-3)=r+1.$

So,

$r=6{k}^{2}-19.$

Now $r$ must be valid for the combinations ${}^{35}{C}_{r}$ and ${}^{36}{C}_{r+1}$.

Thus,

$0\le r\le 35.$

Substitute $r=6{k}^{2}-19$:

$0\le 6{k}^{2}-19\le 35.$

Add $19$ throughout:

$19\le 6{k}^{2}\le 54.$

Divide by $6$:

$\frac{19}{6}\le {k}^{2}\le 9.$

Now possible integer square values are

${k}^{2}=4\, \, {k}^{2}=9.$

So the integer values of $k$ are:

If ${k}^{2}=4$, then $k=\pm 2$

If ${k}^{2}=9$, then $k=\pm 3$

Now find corresponding $r$:

For ${k}^{2}=4$,

$r=6(4)-19=24-19=5.$

So pairs are

$(5,2),(5,-2).$

For ${k}^{2}=9$,

$r=6(9)-19=54-19=35.$

So pairs are

$(35,3),(35,-3).$

Hence total number of elements in the set $S$ is

$4.$

Therefore, the correct answer is:

$4$

So, Option B is correct.

Solutions · 1