Question

Let n and r be integers with 1 ≤ r< n show that

\(C(n,r - 1){\rm{ }} = C(n + 2,r + 1) - 2C(n + 1,r + 1) + C(n,r + 1).\)

Short answer

The given statement is verified:\(C\left( {n + 2,{\rm{ }}r + 1} \right) - C\left( {n + 1,{\rm{ }}r + 1} \right) = C\left( {n,{\rm{ }}r - 1} \right)\).

Step-by-step solution

Definition of integers and Pascal's Identity

An integer is a number from the set of negative and positive numbers, including zero that has no decimal or fractional element.

Pascal's Identity is a useful theorem of combinatorics dealing with combinations

Solution

Let us solve the given problem,

According to Pascal's identity,

\(C(n + 2,r + 1) - C(n + 1,r + 1) = C(n + 1,r)\)

Similarly

\(C(n + 1,r + 1) - C(n,r + 1) = C(n,r)\)

Subtracting the second equation from the first one gives us:

\(C(n + 2,r + 1) - 2C(n + 1,r + 1) + C(n,r + 1) = C(n + 1,r) - C(n,r) = C(n,r - 1)\).

Hence, the given statement is verified.