SATHEE: Permutations And Combinations Question 39

Permutations And Combinations Question 39

Question: A person goes in for an examination in which there are four papers with a maximum of $ m $ marks from each paper. The number of ways in which one can get $ 2m $ marks is

Options:

A) $ ^{2m+3}C_3 $

B) $ \frac{1}{3}(m+1)(2m^{2}+4m+1) $

C) $ \frac{1}{3}(m+1)(2m^{2}+4m+3) $

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

The required number = coefficient of $ x^{2m} $ in $ {{(x^{0}+x^{1}+……+x^{m})}^{4}} $ = coefficient of $ x^{2m} $ in $ {{( \frac{1-{x^{m+1}}}{1-x} )}^{4}} $ = coefficient of $ x^{2m} $ in $ {{(1-{x^{m+1}})}^{4}}{{(1-x)}^{-4}} $ = coefficient of $ x^{2m} $ in $ (1-4{x^{m+1}}+6{x^{2m+2}}+……) $ $ ( 1+4x+……+\frac{(r+1)(r+2)(r+3)}{3\ !}x^{r}+…. ) $ $ =\frac{(2m+1)(2m+2)(2m+3)}{6}-4m\frac{(m+1)(m+2)}{6} $ $ =\frac{(m+1)(2m^{2}+4m+3)}{3} $ .

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Permutations And Combinations Question 39

Question: A person goes in for an examination in which there are four papers with a maximum of $ m $ marks from each paper. The number of ways in which one can get $ 2m $ marks is

Options:

Answer:

Solution:

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