If ( 1^15C0 + 1^15C1) ( 1^15C1 + 1^15C2) ( 1^15C12 + 1^15C13) = ^13^14C0 ^14C1 ^14C12, then 30 is equal to.

Given: ( 1^15C0 + 1^15C1) ( 1^15C1 + 1^15C2) ( 1^15C12 + 1^15C13) = ^13^14C0 ^14C1 ^14C12 Find: 30 Use the identity 1^nCr + 1^nCr+1 = n+1n ^n-1Cr For the general term, Tr = 1^nCr + 1^nCr+1 Then Tr = r!(n-r)!n! + (r+1)!(n-r-1)!n! Tr = r!(n-r-1)!n! [(n-r)+(r+1) ] Tr = r!(n-r-1)!n!(n+1) Tr = n+1n r!(n-r-1)!(n-1)! = n+1n ^n-1Cr Now substitute n=15: 1^15Cr + 1^15Cr+1 = 1615 ^14Cr Hence, r=0^12 ( 1^15Cr + 1^15Cr+1) = r=0^12 1615 ^14Cr Since there are 13 terms, Product = (16/15)^13 r=0^12 ^14Cr = (16/15)^13^14C0 ^14C1 ^14C12 Comparing with the given expression, = 1615 Therefore, 30 = 30 1615 = 32 So, the value of 30 is 32. The solution lists the answer key as 15, but the working shown in the solution gives 32. Therefore, the answer derived from the solution is 32. Given: the product of terms of the form 1^15Cr + 1^15Cr+1 Find: 30 Recognize directly that 1^15Cr + 1^15Cr+1 = 1615 ^14Cr So each factor contributes a common multiplier of 1615 and a denominator factor ^14Cr. Therefore, r=0^12 ( 1^15Cr + 1^15Cr+1) = (16/15)^13^14C0 ^14C1 ^14C12 Hence, = 1615 and so 30 = 32 Therefore, the value of 30 is 32.

If ( 1^15C0 + 1^15C1) ( 1^15C1 + 1^15C2) ( 1^15C12 + 1^15C13) = ^… — Solution | JEE Mathematics

NVAMediumJEE 2026Applications of P&C

If $\left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \dots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 {}^{14}C_1 \dots {}^{14}C_{12}}$ , then $30\alpha$ is equal to _____.

Answer

Correct answer:32

Step-by-step solution

Standard Method

Given:

\left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \dots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 {}^{14}C_1 \dots {}^{14}C_{12}}

Find: $30\alpha$

Use the identity

\frac{1}{{}^{n}C_r} + \frac{1}{{}^{n}C_{r+1}} = \frac{n+1}{n \cdot {}^{n-1}C_r}

For the general term,

T_r = \frac{1}{{}^{n}C_r} + \frac{1}{{}^{n}C_{r+1}}

Then

T_r = \frac{r!(n-r)!}{n!} + \frac{(r+1)!(n-r-1)!}{n!}

T_r = \frac{r!(n-r-1)!}{n!}\left[(n-r)+(r+1)\right]

T_r = \frac{r!(n-r-1)!}{n!}(n+1)

T_r = \frac{n+1}{n} \cdot \frac{r!(n-r-1)!}{(n-1)!} = \frac{n+1}{n \cdot {}^{n-1}C_r}

Now substitute $n=15$ :

\frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}} = \frac{16}{15 \cdot {}^{14}C_r}

Hence,

\prod_{r=0}^{12} \left( \frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}} \right) = \prod_{r=0}^{12} \frac{16}{15 \cdot {}^{14}C_r}

Since there are $13$ terms,

\text{Product} = \frac{(16/15)^{13}}{\prod_{r=0}^{12} {}^{14}C_r} = \frac{(16/15)^{13}}{{}^{14}C_0 {}^{14}C_1 \dots {}^{14}C_{12}}

Comparing with the given expression,

\alpha = \frac{16}{15}

Therefore,

30\alpha = 30 \times \frac{16}{15} = 32

So, the value of $30\alpha$ is $32$ .

The solution lists the answer key as $15$ , but the working shown in the solution gives $32$ . Therefore, the answer derived from the solution is $32$ .

Use the reciprocal binomial identity directly

Given: the product of terms of the form

\frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}}

Find: $30\alpha$

Recognize directly that

\frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}} = \frac{16}{15\cdot {}^{14}C_r}

So each factor contributes a common multiplier of $\frac{16}{15}$ and a denominator factor $^{14}C_r$ . Therefore,

\prod_{r=0}^{12}\left(\frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}}\right) = \frac{(16/15)^{13}}{{}^{14}C_0 {}^{14}C_1 \dots {}^{14}C_{12}}

Hence,

\alpha = \frac{16}{15}

and so

30\alpha = 32

Therefore, the value of $30\alpha$ is $32$ .

Common mistakes

A common mistake is to trust the listed answer key $15$ without checking the working. This is wrong because the algebra in the solution gives $\alpha=\frac{16}{15}$ . Always derive the value from the identity and then compute $30\alpha$ .
Students may count the number of factors incorrectly. The product runs from $r=0$ to $r=12$ , so there are $13$ terms, not $12$ . Use the inclusive count carefully when raising $\frac{16}{15}$ to a power.
Another mistake is to apply the identity incorrectly as $\frac{n+1}{{}^{n-1}C_r}$ and forget the factor of $\frac{1}{n}$ . This changes the value of $\alpha$ . Keep the correct identity as $\frac{n+1}{n\cdot {}^{n-1}C_r}$ .

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