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If ( x + y ) 2 (x+y)^2 has three terms which are x 2 + y 2 + 2 x y x^2+y^2+2xy then how many terms does ( a + b + c + d + + y + z ) 2 (a+b+c+d+\cdots+ y+z)^2 will have?


This problem is original.


The answer is 351.

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1 solution

Sabhrant Sachan
May 7, 2016

We can Do this Question using Logic or by using Properties of Multinomial Thm 1st Method : logic We can see that each term is Basically getting multiplied by 26 letters so Total Terms = 26 × 26 = 676 But only terms like a 2 , b 2 . . z 2 are Unique, Terms like a b , b a and z a , a z will merge Subtracting the unique 26 terms from total terms , we need to divide the rest by 2 We get : Terms = 676 26 2 = 325 Now add the unique terms Back in terms , we get our Total Terms : 325 + 26 = 351 \text{We can Do this Question using Logic or by using Properties of Multinomial Thm } \\ \text{1st Method : logic } \\ \text{We can see that each term is Basically getting multiplied by 26 letters } \\ \text{ so Total Terms = } 26\times26=676 \\ \text{But only terms like }a^2,b^2..z^2 \text{ are Unique, Terms like } ab,ba \text{ and } za ,az\text{ will merge } \\ \text{Subtracting the unique 26 terms from total terms , we need to divide the rest by 2} \\ \text{We get : Terms = }\dfrac{676-26}{2}=325 \\ \text{Now add the unique terms Back in terms , we get our Total Terms : }\boxed{325+26 = 351}

2nd Method : Multinomial From a Property of Binomial Thm. Coefficient of a term in ( a + b + c + d + . . . ) n is : n ! n 1 ! × n 2 ! × n 3 ! × n 4 ! × . . . . × a n 1 × b n 2 × c n 3 × d n 4 × . . . . Where n 1 + n 2 + n 3 + n 4 + = n and 0 n 1 , n 2 , n 3 , n 4 , . . . n In this case n 1 + n 2 + n 3 + + n 26 = 2 Total Number of terms = Number of Integral Solutions of this equation ( 2 + 26 1 26 1 ) ( 27 25 ) Ans : 27 × 26 2 = 351 \text{2nd Method : Multinomial } \\ \text {From a Property of Binomial Thm.} \\ \text{Coefficient of a term in } (a+b+c+d+...)^n \text { is :} \\ \dfrac{n!}{{n_1!}\times{n_2!}\times{n_3!}\times{n_4!}\times{....}}\times{a^{n_1}}\times{b^{n_2}}\times{c^{n_3}}\times{d^{n_4}\times{....}} \\ \text{Where } n_1+n_2+n_3+n_4+\cdots=n \text{ and } 0\le{n_1,n_2,n_3,n_4,...}\le{n}\\ \text{In this case } n_1+n_2+n_3+\cdots+n_{26}=2 \\ \text{Total Number of terms = Number of Integral Solutions of this equation }\\ \implies \binom{2+26-1}{26-1} \implies \binom{27}{25}\\ \implies \text{Ans : }\boxed{\dfrac{27\times{26}}{2}=351}

Very nice :)

Abhay Tiwari - 5 years, 1 month ago

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Try to solve my question

Sabhrant Sachan - 5 years, 1 month ago

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