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asked in Combinatory by Veteran (12.9k points)  
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In multinomial expansion of $(1+x^5+x^9)^{100}$, every term will be of form: $C(100 ; m, n, p) \times 1^m \times (x^5)^n \times (x^9)^p$, where m + n + p =100.

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

So, (m, n, p) will be (97, 1, 2). Now, put these values in the term for $x^{23}$.

C(100; 97, 1, 2) =$\frac{100!}{(97! * 1! * 2!)}$ = 485100. This will be the coefficient of $x^{23}$.

 

answered by Active (2.2k points)  
edited by
in simple way... we can make X^23 using 2 times X^9 +one time X^5 ... SO C(100,2) are the ways to select two X^9 and 98 way for chose X^5

so coefficient of X^23 = C(100,2)*98= 485100
Plz xplain


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