Q37. Show that Cn-1,r+Cn-1,r-1=Cn,r.... [FREE SOLUTION]

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

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

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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91影视

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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