If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5

Question

If  (a+b)^5=

a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5

What is the value of  x+y+z+t ?

A. 7
B. 15
C. 20
D. 30
E. 35

KarishmaB · Accepted Answer

I will just put a = b = 1 here.

(a + b)^5 = (1 + 1)^5 = 2^5 = 32

a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5 = 1 + x + y + z + t + 1 = 32

x + y + z + t = 30

Answer (D)

BrushMyQuant · Answer

(a+b)^5  can be opened using  Binomial Theorem

(a+b)^5 = ^5C_0a^5b^0 + ^5C_1a^4b^1 + ^5C_2a^3b^2 + ^5C_3a^2b^3 + ^5C_4a^1b^4 + ^5C_5a^0b^5

So, x + y + z + t =  ^5C_1 + ^5C_2 + ^5C_3 + ^5C_4 
=  \frac{5!}{1!*4!} + \frac{5!}{2!*3!} + \frac{5!}{3!*2!} + \frac{5!}{4!*1!} 
=  5 + \frac{5*4}{2} + \frac{5*4}{2 }+ 5  = 30

So,  Answer will be D 
Hope it helps!

Check out the below Video to MASTER Binomial Theorem

https://www.youtube.com/watch?v=h1HTEhmUb1w

Kinshook · Answer

Given: If  (a+b)^5=

a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5

Asked: What is the value of  x+y+z+t ?

(a+b)^5=a^5+^5C_1a^4b+^5C_2a^3b^2+^5C_3a^2b^3+^5C_4ab^4+b^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5C4ab^4+b^5

x = 5; y=z=10; t = 5
x + y + z + t = 5 + 10 + 10 + 5 = 30

IMO D

rajdeep212003 · Answer

This is a good one short to do but nice and tricky

If (a+b)^5 = a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5

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