If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.
Question:

 If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.

[Hint: From equation using nCnCr + 1 and nCnCr – 1 to find the value of r.]

Solution:

We know that,

nCr

$=\frac{n !}{r !(n-r) !}$

According to the question,

nCr – 1 =36,

nC=84,

 

nCr +1=126

$\frac{C_{r}^{n}}{C_{r+1}^{n}}=\frac{84}{126}$

$\Rightarrow \frac{\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}}{\frac{\mathrm{n} !}{(\mathrm{r}+1) !(\mathrm{n}-\mathrm{r}-1) !}}=\frac{84}{126}=\frac{2}{3}$

2n-2r=3r+3

⇒2n – 3 = 5r … (i)

$\frac{C_{r}^{n}}{C_{r-1}^{n}}=\frac{84}{36}$

$\Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r-1) !(n-r+1) !}}=\frac{7}{3}$

=3n-3r+3=7r

3n+3=10r … (ii)

From (i) and (ii),

We get,

2(2n – 3) = 3n+3

4n – 3n – 6 – 3=0

n=9

And r=3

Now

rC2=3C2 = 3!/2!

=3

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