The number of ordered pairs (r, k) for which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot ^{36}C_{r+1}$.Using the identity $^{n}C_{r} = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ or $^{n

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the equation: $6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot ^{36}C_{r+1}$.Using the identity $^{n}C_{r} = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ or $^{n+1}C_{r+1} = \frac{n+1}{r+1} \cdot ^{n}C_{r}$,we have $^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_{r}$.Substituting this into the equation:$6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot \frac{36}{r+1} \cdot ^{35}C_{r}$.Since $^{35}C_{r} 
eq 0$,we can divide both sides by $^{35}C_{r}$:$6 = (k^{2} - 3) \cdot \frac{36}{r+1}$.Rearranging for $(k^{2} - 3)$:$k^{2} - 3 = \frac{6(r+1)}{36} = \frac{r+1}{6}$.Thus,$k^{2} = \frac{r+1}{6} + 3 = \frac{r+19}{6}$.Since $k$ is an integer,$k^{2}$ must be a perfect square. Also,$0 \le r \le 35$.For $r=5$,$k^{2} = \frac{5+19}{6} = 4 \Rightarrow k = \pm 2$.For $r=35$,$k^{2} = \frac{35+19}{6} = 9 \Rightarrow k = \pm 3$.These give the ordered pairs $(5, 2), (5, -2), (35, 3), (35, -3)$.There are $4$ such ordered pairs.

The number of ordered pairs $(r, k)$ for which $6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot ^{36}C_{r+1}$,where $k$ is an integer,is

Similar Questions

There are two urns. Urn $A$ has $3$ distinct red balls and urn $B$ has $9$ distinct blue balls. From each urn,two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

Out of $6$ books,in how many ways can a set of one or more books be chosen?

$A$ committee of $7$ members is to be formed from $6$ boys and $4$ girls such that the number of boys is greater than the number of girls. In how many ways can this be done?

$A$ student has to answer a multiple-choice question having $5$ alternatives in which two or more than two alternatives are correct. Then the number of ways in which the student can answer that question is

Vedclass Test Series

Exam Paper Generator

Online Exam Module