JEE MATHEMATICS MCQ: COMBINATIONS

1 Easy

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Application +4 | -1

\sum\limits_{k = 1}^{31} {\left( {^{31}{C_k}} \right)} \left( {^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {^{30}{C_k}} \right)} \left( {^{30}{C_{k - 1}}} \right) = \frac{{\alpha \left( {60!} \right)}}{{\left( {30!} \right)\left( {31!} \right)}},

where

\alpha \in R,

then the value of

16\alpha

is equal to

View answer and explanation

Correct option: A

1411

Option A (1411) is correct because it is obtained by simplifying the summation expression using combinatorial identities and then solving for α.

More Questions from COMBINATIONS

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m \times n

m

n

A

B

{\;^{n + 2}}{C_8}:{\;^{n - 2}}{P_4} = \frac{{57}}{{16}}

n