Permutations and Combinations 2 Question 18
18. The product of any $r$ consecutive natural numbers is always divisible by $r$ !.
(1985, 1M)
Show Answer
Answer:
Correct Answer: 18. (True)
Solution:
- Let $r$ consecutive integers be $x+1, x+2, \ldots, x+r$.
$ \begin{aligned} \therefore(x+1)(x+2) \ldots(x+r) & =\frac{(x+r)(x+r-1) \ldots(x+1) x !}{x !} \\ & =\frac{(x+r) !}{(x) !} \cdot \frac{r !}{r !}={ }^{x+r} C _r \cdot(r) ! \end{aligned} $
Thus, $(x+1)(x+2) \ldots(x+r)={ }^{x+r} C _r \cdot(r) !$, which is clearly divisible by $(r) !$. Hence, it is a true statement.
