SATHEE: Permutations And Combinations Question 161

Permutations And Combinations Question 161

Question: Out of 5 apples, 10 mangoes and 15 oranges, any 15 fruits distributed among two persons. The total number of ways of distribution [DCE 2005]

Options:

A) 66

B) 36

C) 60

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

Number of ways = coefficient of $ x^{15} $ in the expansion $ (1+x+x^{2}+x^{3}+x^{4}+x^{5}) $ $ (1+x+x^{2}+…….+x^{10}) $ $ (1+x+x^{2}+……+x^{15}) $ $ (1+x+x^{2}+x^{3}+x^{4}+x^{5})(1+x+x^{2}+…..+x^{10}) $ $ (1+x+x^{2}+…+x^{15})=(1-x^{6}-x^{11})(1+{{}^{3}}C_1x+{{}^{4}}C_2x^{2} $ $ +……+{{}^{6}}C_4x^{4}+{{}^{11}}C_9x^{9}+{{}^{17}}C _{15}x^{15}+…………) $ $ =…….+…….+x^{15}({{-}^{11}}C_9-{{}^{6}}C_4+{{}^{17}}C _{15}) $ $ =…….+……+x^{15}(-55-15+136) $ $ =x^{15}\times 66 $ \ Coefficient of $ x^{15}=66 $ .

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Permutations And Combinations Question 161

Question: Out of 5 apples, 10 mangoes and 15 oranges, any 15 fruits distributed among two persons. The total number of ways of distribution [DCE 2005]

Options:

Answer:

Solution:

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