If
\(\sum\limits_{k=1}^{31}\) \((^{31}C_k) (^{31}C_{k-1})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) (^{30}C_{k-1})\) \(= \frac{α (60!)} {(30!) (31!)}\)
where \(α ∈ R\), then the value of 16α is equal to
If
\(\sum\limits_{k=1}^{31}\) \((^{31}C_k) (^{31}C_{k-1})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) (^{30}C_{k-1})\) \(= \frac{α (60!)} {(30!) (31!)}\)
where \(α ∈ R\), then the value of 16α is equal to
- 1411
- 1320
- 1615
- 1855
The Correct Option is A
Solution and Explanation
The correct answer is (A) : 1411
\(\sum\limits_{k=1}^{31}\)\((^{31}C_k) (^{31}C_{k-1})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) (^{30}C_{k-1})\)
\(=\sum\limits_{k=1}^{31}\) \((^{31}C_k) . (^{31}C_{32-k})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) . (^{30}C_{k-1})\)
\(= ^{62}C_{32} - ^{60}C_{31}\)
\(= \frac{60!}{31!29!} ( \frac{62.61}{32.30} - 1 ) = \frac{60!}{ 31!29!} \frac{2822}{32.30}\)
\(α = \frac{2822}{32}\)
\(⇒ 16α = 1411\)
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Concepts Used:
Permutations and Combinations
Permutation:
Permutation is the method or the act of arranging members of a set into an order or a sequence.
- In the process of rearranging the numbers, subsets of sets are created to determine all possible arrangement sequences of a single data point.
- A permutation is used in many events of daily life. It is used for a list of data where the data order matters.
Combination:
Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.
- Combination refers to the combination of about n things taken k at a time without any repetition.
- The combination is used for a group of data where the order of data does not matter.