JEE Novice - (7)

Probability Level 2

$\Large {n \choose r-1} : {n \choose r} : {n \choose r+1} = 2:4:5$

If positive integers $n,r$ satisfy the ratio above , find the value of $n+r$ .

This question is a part of JEE Novices .

The answer is 11.

1 solution

Curtis Clement
May 26, 2015

Using the LHS: $\frac{2 n!}{(r-1)!(n-r+1)!} = \frac{n!}{r!(n-r)!}$ Cancelling n! and cross - multiplying: $2 r! (n-r)! = (r-1)! (n-r+1)!$ Now using the property that r! = r(r-1)! etc... $\ 2r = n-r+1 \Rightarrow\ 3r = n+1 \ .......... (1)$

Using the RHS: $\frac{5 n!}{r!(n-r)!} = \frac{4 n!}{(r+1)!(n-r-1)!}$ $5(r+1)! (n-r-1)! = 4(r!)(n-r)! \Rightarrow\ 5(r+1) = 4(n-r)$ $9r +5 = 4n \ .......... (2)$ Substituting (1) into (2): $3n+8 = 4n \Rightarrow\ n=8 \Rightarrow\ r = 3$ $\therefore\ n+r = 11$

Good job Curtis!I solved it quite the same way but I also noticed that 2+4+5=11 is there an explanation for that?

Arian Tashakkor - 6 years ago

Nope, my guess is that it's just coincidental because if the ratio was $1:2:3$ , then we get $n=14$ and $r=5$ which gives the answer as $19$ which is obviously $\neq 1+2+3=6$ .

Prasun Biswas - 6 years ago

JEE Novice - (7)

This question is a part of JEE Novices .

The answer is 11.

1 solution

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