Q37.

Question

Show that $C (n - 1, r) + C (n - 1, r - 1) = C (n, r)$ .

Step-by-Step Solution

Verified

Answer

It is Proved that $C (n - 1, r) + C (n - 1, r - 1) = C (n, r)$ .

1Step 1. Given Information.

Given to prove that $C (n - 1, r) + C (n - 1, r - 1) = C (n, r)$ .

2Step 2. Calculation .

When n objects are available and r are to be picked then the number of combinations is given by $C (n, r) = \frac{n!}{(n - r)! r!}$ .

Consider, $C (n - 1, r) + C (n - 1, r - 1)$

$\begin{array}{l} C (n - 1, r) + C (n - 1, r - 1) = \frac{(n - 1)!}{(n - 1 - r)! r!} + \frac{(n - 1)!}{((n - 1) - (r - 1))! (r - 1)!} \\ C (n - 1, r) + C (n - 1, r - 1) = \frac{(n - 1)!}{(n - r - 1)! r!} + \frac{(n - 1)!}{(n - 1 - r + 1)! (r - 1)!} \\ C (n - 1, r) + C (n - 1, r - 1) = \frac{(n - 1)! \cdot (n - r)}{(n - r - 1)! r! \cdot (n - r)} + \frac{(n - 1)! \cdot r}{(n - r)! (r - 1)! \cdot r} \\ C (n - 1, r) + C (n - 1, r - 1) = \frac{(n - 1)! \cdot (n - r)}{(n - r)! r!} + \frac{(n - 1)! \cdot r}{(n - r)! r!} \\ C (n - 1, r) + C (n - 1, r - 1) = \frac{(n - 1)!}{(n - r)! r!} [n - r + r] \\ C (n - 1, r) + C (n - 1, r - 1) = \frac{(n - 1)! \cdot n}{(n - r)! r!} \\ C (n - 1, r) + C (n - 1, r - 1) = \frac{n!}{(n - r)! r!} \\ C (n - 1, r) + C (n - 1, r - 1) = C (n, r) \end{array}$

Therefore, from the values of the expressions, $C (n - 1, r) + C (n - 1, r - 1) = C (n, r)$ .

3Step 3. Conclusion .

Therefore, It is Proved that $C (n - 1, r) + C (n - 1, r - 1) = C (n, r)$ .

Q36.

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