Combinatorics..#1

The coefficient of a b c 3 d e 2 abc^{3}de^{2} in the expansion of ( a + b + c + d + e ) 8 (a+b+c+d+e)^{8} is equal to

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3600 3360 None 3630

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1 solution

Melissa Quail
Feb 18, 2015

Using the multinominal theorem, the multinominal coefficients are n ! k 1 ! k 2 ! k 3 ! . . . . . k m \frac{n!}{k_1!k_2!k_3!.....k_m} where n is the power to which the original expression was raised and k 1 + k 2 + k 3 . . . . . k m = n k_1+k_2+k_3.....k_m= n .

So we want to work out the coefficient of a b c 3 d e 2 abc^3de^2 = a 1 b 1 c 3 d 1 e 2 a^1b^1c^3d^1e^2 . In this case n = 8 and k 1 , k 2 , k 3 . . . k m k_1, k_2, k_3...k_m = 1, 1, 3, 1, 2 (the powers in the expression a b c 3 d e 2 abc^3de^2 ).

This means that the coefficient is 8 ! 1 ! 1 ! 3 ! 1 ! 2 ! \frac{8!}{1! 1! 3! 1! 2!} = 3360 \boxed{3360} .

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