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D
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Difficulty:
65%
(hard)
Question Stats:
59% (02:05) correct
41%
(01:58)
wrong
based on 270
sessions
History
Date
Time
Result
Not Attempted Yet
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
\(a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\)
What is the value of \(x+y+z+t\)?
A. 7
B. 15
C. 20
D. 30
E. 35
28 Nov 2016, 10:28
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nguyendinhtuong
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)
General Discussion
Updated on: 30 May 2025, 18:16
Originally posted by BrushMyQuant on 28 Nov 2016, 07:32.
Last edited by BrushMyQuant on 30 May 2025, 18:16, edited 6 times in total.
Last edited by BrushMyQuant on 30 May 2025, 18:16, edited 6 times in total.
Kudos
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\((a+b)^5\) 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
\((a+b)^5 = ^5C_0a^5b^0 + ^5C_1a^4b^1 + ^5C_2a^3b^2 + ^5C_3a^2b^3 + ^5C_4a^1b^4 + ^5C_5a^0b^5\)
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
