SATHEE: Permutations And Combinations Question 367

Permutations And Combinations Question 367

Question: If $ \frac{2}{9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{2^{a}}{b!}, $ where $ a,b\in N, $ then the ordered pair (a, b) is

Options:

A) $ (9,10) $

B) $ (10,9) $

C) $ (7,10) $

D) $ (10,7) $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $ \frac{2}{9!}+\frac{2}{3!7!}+\frac{1}{5!5!} $ $ =\frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{3!7!}+\frac{1}{9!1!} $ $ =\frac{1}{10!}{{{}^{10}}C_1+{{}^{10}}C_3+{{}^{10}}C_5+{{}^{10}}C_7+{{}^{10}}C_9} $ $ =\frac{1}{10!}({2^{10-1}})=\frac{2^{9}}{10!}=\frac{2^{a}}{b!} $ (given)
$ \Rightarrow a=9,b=10 $

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Permutations And Combinations Question 367

Question: If $ \frac{2}{9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{2^{a}}{b!}, $ where $ a,b\in N, $ then the ordered pair (a, b) is

Options:

Answer:

Solution:

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