Answer to Question #301937 in Discrete Mathematics for Tege

Question #301937

9. In the expression (1+x5+x9)10 find the coefficient of a) x23 b) x32

1
Expert's answer
2022-02-24T15:40:44-0500

In multinomial expansion of (1+x5+x9)10,(1+x^5+x^9)^{10}, every term will be of form:


C(10;m,n,p)×1m×(x5)n×(x9)p,C(10; m, n, p)\times1^m\times(x^5)^n\times (x^9)^p,

where m+n+p=10.m + n + p =10.

a) For the coefficient of x23,x^{23}, (5n+9p)(5n + 9p) should be equal to 23.23. There is only one pair exist for this condition to hold i.e (1,2).(1,2).

(m,n,p)(m,n,p) will be (7,1,2).(7, 1, 2).


C(10;7,1,2)=10!7!1!2!=10(9)(8)2=360C(10; 7, 1, 2)=\dfrac{10!}{7!1!2!}=\dfrac{10(9)(8)}{2}=360

The coefficient of x23x^{23} is 360.360.


b) For the coefficient of x32,x^{32}, (5n+9p)(5n + 9p) should be equal to 32.32. There is only one pair exist for this condition to hold i.e (1,3).(1,3).

(m,n,p)(m,n,p) will be (6,1,3).(6, 1, 3).

C(10;6,1,3)=10!6!1!3!=10(9)(8)(7)6=840C(10; 6, 1, 3)=\dfrac{10!}{6!1!3!}=\dfrac{10(9)(8)(7)}{6}=840

The coefficient of x32x^{32} is 840.840.



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