Question

Determine the number of vectors $(x_1,...,x_n),$such that each $x_i$is either $0$or$1$and$\sum _{i=1}^n{x}_i\ge k$.

Short answer

There are $\sum _{m=k}^n\frac{n!}{m!(n-m)!}$different ways.

Step-by-step solution

Given Information.

The number of vectors $\left(x_1,\dots ,x_n\right)$is either 0 or 1and$\sum _{i=1}^n{x}_i\ge k$.

Explanation.

The condition $\sum _{i=1}^n{x}_i\ge k$holds if and only if the vector $x=\left(x_1,\dots ,x_n\right)$has $k$or more components equal to$1$.

Let us fix that the vector shall have exactly $m$components equal to$1$. Therefore, we can arrange this collection of ones and zeros in $n!$different ways. But, because we do not want to count twice the permutations of zeros and ones among ourselves, we should divide byrole="math" localid="1648738905041" $m!(m-k)!$

The result follows by considering all $m$between $k$and$n$.

Explanation.

There are $\sum _{m=k}^n\frac{n!}{m!(n-m)!}$different ways.