Question

The number of ordered pairs (r , k) for which $6{}^{35}{C_r} = \left( {{k^2} - 3} \right){}^{36}{C_{r + 1}}$, where k is an integer is,
A.6
B.2
C.4
D.3

Answer 1

Use the expansion of ${}^n{C_r} = \dfrac{{{}^n{P_r}}}{{\left( {n - r} \right)!}}$ . and on expanding on both the sides , cancel out the common terms from both the sides and calculate up to the final relation between r and k . And thus from all the available values of r and k we can form all the possible ordered pairs of r and k .

Complete step-by-step answer:
Now, expand both side of $6{}^{35}{C_r} = \left( {{k^2} - 3} \right){}^{36}{C_{r + 1}}$ by the formula stated in the hint.
$6\dfrac{{35!}}{{r!\left( {35 - r!} \right)}} = \left( {{k^2} - 3} \right)\dfrac{{36!}}{{\left( {r + 1} \right)!\left( {36 - \left( {r + 1} \right)} \right)!}}$
Also use the property that, $\begin{gathered} r! = r.(r - 1)! \\\ r! = r.(r - 1).(r - 2)........3.2.1 \\\ \end{gathered}$
$6\dfrac{{35!}}{{r!\left( {35 - r!} \right)}} = \left( {{k^2} - 3} \right)\dfrac{{36!}}{{\left( {r + 1} \right)!\left( {35 - r} \right)!}}$
Further simplifying the above equation,
$6\dfrac{1}{{r!}} = \left( {{k^2} - 3} \right)\dfrac{{36}}{{\left( {r + 1} \right)r!}}$
$6 = \dfrac{{36\left( {{k^2} - 3} \right)}}{{\left( {r + 1} \right)}}$
$\dfrac{1}{6} = \dfrac{{\left( {{k^2} - 3} \right)}}{{\left( {r + 1} \right)}}$
$\dfrac{{r + 1}}{6} = {k^2} - 3$
${k^2} = 3 + \dfrac{{r + 1}}{6} = \dfrac{{r + 19}}{6}$ …(1)
Now, as k is an integer stated in the question,
Thus, $\dfrac{{r + 19}}{6}$is a perfect square
Now, $6{}^{35}{C_r} = \left( {{k^2} - 3} \right){}^{36}{C_{r + 1}}$ will exist only if
As we know from the above given equation in the question and in general the value of r can be less than or equal to n and so r varies as stated below.
$0 \leqslant r \leqslant 35$
$19 \leqslant r + 19 \leqslant 54$
$\dfrac{{19}}{6} \leqslant \dfrac{{r + 19}}{6} \leqslant 6$
$\dfrac{{19}}{6} \leqslant {k^2} \leqslant 6$
$4 \leqslant {k^2} \leqslant 6$
Thus, for, $4 \leqslant {k^2} \leqslant 6$
For k to be an integer,
${k^2} = 4$
$k = \pm 2$ …(2)
$r + 1 = 6$
$r = 5$ … (3)
Hence the ordered pair (r , k) is (5,2) and (5,-2)
Two ordered pairs are possible and option (b) is the correct answer.

Note: Combination is selection of all or part of sets of objects, regardless of the order in which objects are selected, whereas in permutation the order of selection matters.
Also, ${}^n{C_r} = \dfrac{{{}^n{P_r}}}{{\left( {n - r} \right)!}}$