Question

Show that, if n is a positive integer, then \(\sum\limits_{j = 1}^n C (n,j) = n!\).

Short answer

It is true for all positive integers, n that \(\sum\limits_{j = 1}^n C (n,j) = n!\).

Step-by-step solution

 Given

The given result is \(\sum\limits_{j = 1}^n C (n,j) = n!\)

The Concept of moment and forces

The objects are labelled but the boxes are unlabelled, we can use the Stirling numbers formula

\(\sum\limits_{j = 1}^k {\frac{1}{{j!}}} \sum\limits_{i = 0}^{j - 1} {{{( - 1)}^i}} C(ji){(j - i)^n}\)

Prove of result

The given result is \(\sum\limits_{j = 1}^n C (n,j) = n!\)

Let us prove the given result by mathematical induction.

Step 1: put\(n = 1\)in the given result, we get

\(\begin{array}{c}\sum\limits_{j = 1}^1 C (n,j) = 1!\\C(1,1) = 1\end{array}\)

The given result is true for true\(n = 1\).

Step 2: assume the result true for\(n = m\)

\(\sum\limits_{j = 1}^m C (m,j) = m!\forall m \ge 1\)

Step 3: We shall prove the given result true for \(n = m + 1\)

That is \(\sum\limits_{j = 1}^{m + 1} C (m + 1,j) = (m + 1)!\forall m \ge 1\)

Consider left hand side of above equation, we get

\(\begin{array}{c}\sum\limits_{j = 1}^{m + 1} C (m + 1,j) = \sum\limits_{j = 1}^m C (m,j) + k!(k + 1)\\ = k! + k!(k + 1)\\ = k!(k + 1)\end{array}\)

\(\sum\limits_{j = 1}^{m + 1} C (m + 1,j) = (k + 1)!\)

Therefore, the given result true for \(n = m + 1\),

Thus, by principal of mathematical induction, the given result is true for all positive integers.

Hence, \(\sum\limits_{j = 1}^n C (n,j) = n!\)