Latest TCS Aptitude Question SOLUTION: 3.What is k=28 ∑k^2(28Ck) k=0 , where 28CK is the number of ways of choosing k items from 28 items? a.406 *2^27 b.306 *2^26 c.28 *2^27 d.56 *2^27

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(#M40019974) TCS Logical Reasoning question Keep an EYE

3.What is
k=28
∑k^2(28Ck)
k=0
, where 28CK
is the number of ways of choosing k items from 28
items?
a.406 *2^27
b.306 *2^26
c.28 *2^27
d.56 *2^27

Asked In TCS krishna (10 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Tejaswi  (10 years ago)

dear admin.......plz explain the solution

Is this solution Helpfull? Yes (41) | No (7)  

Abhinav Gupta  (9 years ago)

∑ k^2 C(n,k) = ∑ {k(k-1)+k} C(n,k)
= ∑ k(k-1) C(n,k) + ∑ k C(n,k)
= ∑ k(k-1) {n(n-1) /k(k-1)} C(n-2,k-2) + ∑ k {n/k) C(n-1,k-1)
= ∑ n(n-1) C(n-2,k-2) + ∑ n C(n-1,k-1)
= n(n-1) ∑ C(n-2,k-2) + n ∑ C(n-1,k-1)
= n(n-1) 2^n-2 + n 2^n-1
Now put n=28 , so the answer is
28 * 27 * 2^26 + 28 * 2^27
= 2^27 (14*27 +28)
= 406 * 2^27

Is this solution Helpfull? Yes (17) | No (4)  

Akshay Siddhpura  (10 years ago)

@GEEVEE : please explain how this formula comes and in which case this can apply??

Is this solution Helpfull? Yes (10) | No (4)  

GEEVEE  (10 years ago)

ans:(1/2)*n(n+1)2^(n-1)..
option a
is the correct one

Is this solution Helpfull? Yes (8) | No (17)  

Pavani  (9 years ago)

Consider (1+x)n=C0+C1x+C2x2+…..+Cnxn …….(1)
Differentiating w.r.t x we get
n(1+x)n−1=C1+2C2x+3C3x2+…..+nCnxn−1
Multiplying by x on both sides,
x.n(1+x)n−1=x.C1+2C2x2+3C3x3+…..
Now again differentiating w.r.t to x
n(1+x)n−1+n(n−1)x(1+x)n−2=C0+22C1x+32C2x2+42C3x3…..
Putting x = 1, we get
n(n+1)2n−2=C1+22C2+32C3+42C4
Now substituting n = 28
28(28+1)228−2 = 812.226 = 406.227

Is this solution Helpfull? Yes (3) | No (4)  

ZP  (9 years ago)

Use the formula and get the answer 0.5* n*(n+1)*2^(n-1) ..................put n= 28

.... A) 406* 2^ 27

Is this solution Helpfull? Yes (3) | No (4)  

Ashish  (10 years ago)

a.406*2^27

Is this solution Helpfull? Yes (2) | No (7)  

pragya mehrotra  (9 years ago)

plz explain it ...

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