SATHEE: Permutations and Combinations 2 Question 18

Permutations and Combinations 2 Question 18

18. The product of any $r$ consecutive natural numbers is always divisible by $r$ !.

(1985, 1M)

Analytical & Descriptive Questions

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Answer:

Correct Answer: 18. (True)

Solution:

Let $r$ consecutive integers be $x + 1, x + 2, \dots, x + r$ .

$\begin{aligned} ∴ (x + 1) (x + 2) \dots (x + r) & = \frac{(x + r) (x + r - 1) \dots (x + 1) x!}{x!} \\ = \frac{(x + r)!}{(x)!} \cdot \frac{r!}{r!} =^{x + r} C_{r} \cdot (r)! \end{aligned}$

Thus, $(x + 1) (x + 2) \dots (x + r) =^{x + r} C_{r} \cdot (r)!$ , which is clearly divisible by $(r)!$ . Hence, it is a true statement.

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Permutations and Combinations 2 Question 18

18. The product of any r consecutive natural numbers is always divisible by r !.

Analytical & Descriptive Questions

Answer:

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18. The product of any $r$ consecutive natural numbers is always divisible by $r$ !.