Chapter 12: Q37. (page 642)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

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

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

Step 3. Conclusion.

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

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

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Conclusion.

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Show that Cn-1,r+Cn-1,r-1=Cn,r.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Conclusion.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

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