TCS Company Numerical Ability Permutation and Combination

What is∑ 28 k=0 k^2(28Ck), where 28Ck is the number of ways of choosing k items from 28 items?

Read Solution (Total 1)

∑ 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
7 years agoHelpfull: Yes(1) No(1)

TCS Other Question

A permutation is often represented by the cycles it has. For example, if we permute the numbers in the natural order to 2 3 1 5 4, this is represented as (1 3 2) (5 4). In this the (132) says that the first number has gone to the position 3, the third number has gone to the position 2, and the second number has gone to position 1, and (5 4) means that the fifth number has gone to position 4 and the fourth number has gone to position 5. The numbers with brackets are to be read cyclically.
If a number has not changed position, it is kept as a single cycle. Thus 5 2 1 3 4 is represented as (1345)(2).
We may apply permutations on itself If we apply the permutation (132)(54) once, we get 2 3 1
5 4. If we apply it again, we get 3 1 2 4 5 , or (1 2 3)(4) (5)
If we consider the permutation of 7 numbers (1457)(263), what is its order (how many times must it be applied before the numbers appear in their original order)? In a deck of 52 cards, how many ways are there to select 13 Spade and 13 heart cards without repetition