197 views
1 votes
1 votes

Give a combinatorial argument to establish the identity below for any nonnegative integer n:

$\sum_{k=0}^{n} k\binom{n}{k}=n\ast 2^{n-1}$

1 Answer

1 votes
1 votes
> \begin{align*}  &A = \sum_{k\geq1}^{n}k\cdot \binom{n}{k} \\ \end{align*}

$A $ is the no of ways to form a committee of $k \geq 1$ people out of $n$ available individuals and select one head for the selected committee.

Empty committee is not possible. Therefore we select the head first in $n$ ways and then any subset out of $(n-1)$ remaining people in $2^{n-1} $ ways.

> \begin{align*}  &A =n\cdot2^{n-1} \\  \end{align*}

Related questions

0 votes
0 votes
0 answers
1
Swarnava Bose asked Jun 8, 2023
224 views
What is the total number of integer partitions ( unordered Summation) of the natural number 8 ?I am getting 22. Is it correct ?
0 votes
0 votes
1 answer
2
Swarnava Bose asked Jun 5, 2023
246 views
Given there are 3 full baskets of apples, mangoes, and oranges. How many ways possible ifa) You need to buy any 4 fruits out of these 3 baskets ?b) you buy any 4 fruits s...
0 votes
0 votes
1 answer
3
Swarnava Bose asked Jun 3, 2023
423 views
A power series expression has been converted to Partial Fractions to get :-$\frac{3}{1+5x} - \frac{2}{7-2x}+ \frac{5x}{3+2x} + \frac{7x}{5-2x}$Find the Coefficient of $x^...
22 votes
22 votes
1 answer
4
P C asked Dec 31, 2022
1,621 views
What is the recurrence relation for the ternary strings of length $n$ which can be constructed using 0,1 or 2 only such that the number of 0’s and number of 1's is od...