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MCQMediumJEE 2024Combinations (C(n,r))

JEE Mathematics 2024 Question with Solution

If n+1Cr+1:nCr:n−1Cr−1=55:35:21^{n+1}C_{r+1} : ^nC_r : ^{n-1}C_{r-1} = 55 : 35 : 21, then 2n+5r2n + 5r is equal to:

  • A

    6060

  • B

    6262

  • C

    5050

  • D

    5555

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: n+1Cr+1:nCr:n−1Cr−1=55:35:21^{n+1}C_{r+1} : ^nC_r : ^{n-1}C_{r-1} = 55 : 35 : 21

Find: 2n+5r2n + 5r

From the given ratio,

nCrn−1Cr−1=3521=53\frac{^nC_r}{^{n-1}C_{r-1}} = \frac{35}{21} = \frac{5}{3}

and

n+1Cr+1nCr=5535=117\frac{^{n+1}C_{r+1}}{^nC_r} = \frac{55}{35} = \frac{11}{7}

Using factorial form,

nCrn−1Cr−1=n!r!(n−r)!×(r−1)!(n−r)!(n−1)!=nr\frac{^nC_r}{^{n-1}C_{r-1}} = \frac{n!}{r!(n-r)!} \times \frac{(r-1)!(n-r)!}{(n-1)!} = \frac{n}{r}

So,

nr=53\frac{n}{r} = \frac{5}{3}

which gives

3n=5r3n = 5r

Also,

n+1Cr+1nCr=(n+1)!(r+1)!(n−r)!×r!(n−r)!n!=n+1r+1\frac{^{n+1}C_{r+1}}{^nC_r} = \frac{(n+1)!}{(r+1)!(n-r)!} \times \frac{r!(n-r)!}{n!} = \frac{n+1}{r+1}

Hence,

n+1r+1=117\frac{n+1}{r+1} = \frac{11}{7}

so

7(n+1)=11(r+1)7(n+1) = 11(r+1)

which simplifies to

7n−11r=47n - 11r = 4

From 3n=5r3n = 5r, we get

n=5r3n = \frac{5r}{3}

Substitute into 7n−11r=47n - 11r = 4:

7(5r3)−11r=47\left(\frac{5r}{3}\right) - 11r = 4 35r3−11r=4\frac{35r}{3} - 11r = 4 35r−33r=1235r - 33r = 12 2r=122r = 12 r=6r = 6

Then,

n=53×6=10n = \frac{5}{3} \times 6 = 10

Now,

2n+5r=2(10)+5(6)=20+30=502n + 5r = 2(10) + 5(6) = 20 + 30 = 50

Therefore, the correct option is C.

Direct Ratio Method

Given: n+1Cr+1:nCr:n−1Cr−1=55:35:21^{n+1}C_{r+1} : ^nC_r : ^{n-1}C_{r-1} = 55 : 35 : 21

Find: 2n+5r2n + 5r

Use the direct identities

nCrn−1Cr−1=nr\frac{^nC_r}{^{n-1}C_{r-1}} = \frac{n}{r}

and

n+1Cr+1nCr=n+1r+1\frac{^{n+1}C_{r+1}}{^nC_r} = \frac{n+1}{r+1}

These work because the factorial terms cancel immediately.

So,

nr=3521=53\frac{n}{r} = \frac{35}{21} = \frac{5}{3}

and

n+1r+1=5535=117\frac{n+1}{r+1} = \frac{55}{35} = \frac{11}{7}

From the first, 3n=5r3n = 5r. Solving together with

7(n+1)=11(r+1)7(n+1) = 11(r+1)

gives

r=6,n=10r = 6, \quad n = 10

Hence,

2n+5r=2(10)+5(6)=502n + 5r = 2(10) + 5(6) = 50

Therefore, the correct option is C.

Common mistakes

  • Using nCrn−1Cr−1\frac{^nC_r}{^{n-1}C_{r-1}} incorrectly as n−r+1r\frac{n-r+1}{r}. That formula applies to coefficients with the same upper index. Here both indices change, so use factorial cancellation to get nr\frac{n}{r}.

  • Writing n+1Cr+1nCr=n+1r\frac{^{n+1}C_{r+1}}{^nC_r} = \frac{n+1}{r}. This misses the shift in the lower index. The correct simplification is n+1r+1\frac{n+1}{r+1}.

  • Solving the two linear relations inconsistently after obtaining 3n=5r3n = 5r and 7n−11r=47n - 11r = 4. First express one variable correctly, then substitute carefully to avoid losing the integer values of nn and rr.

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