Question

If $^{36}C_{r+1} = 6 \cdot \frac{^{35}C_r}{k^2 - 3}$, then number of ordered pairs $(r, k)$, where $r, k \in Z$ is ______.

Answer 1

Step 1: Simplify the combination equation.
We know the identity: $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$.
Applying it to $^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_r$.
Substitute this into the given equation:
$\frac{36}{r+1} \cdot ^{35}C_r = \frac{6 \cdot ^{35}C_r}{k^2 - 3}$
$\frac{6}{r+1} = \frac{1}{k^2 - 3} \implies k^2 - 3 = \frac{r+1}{6} \implies k^2 = \frac{r+19}{6}$


Step 2: Determine range for $r$.
For $^nC_r$ to be defined, $0 \le r \le n$.
So for $^{36}C_{r+1}$, $0 \le r+1 \le 36 \implies -1 \le r \le 35$.
Also, since $r, k$ are integers, $r+19$ must be a multiple of 6 and $(r+19)/6$ must be a perfect square.


Step 3: Test values.
If $k^2 = 1$, then $r+19 = 6 \implies r = -13$ (Not in range).
If $k^2 = 4$, then $r+19 = 24 \implies r = 5$. (Valid: $r=5, k = \pm 2$).
If $k^2 = 9$, then $r+19 = 54 \implies r = 35$. (Valid: $r=35, k = \pm 3$).
If $k^2 = 16$, then $r+19 = 96 \implies r = 77$ (Out of range).


Step 4: Count ordered pairs.
Pairs $(r, k)$ are: $(5, 2), (5, -2), (35, 3), (35, -3)$.
Total number of pairs = 4.