Question

State whether the given statement is true or false.
Product of r consecutive integers is divisible by r!.

Answer 1

Hint: Let us use the combination ${}^{n + r}{C_r}$ and expand that to find whether the statement is true or false.

Complete step-by-step answer:
As we know that consecutive numbers are the numbers which are one after other or we can also say that two consecutive numbers have a difference of 1.
Like the set of some consecutive numbers will be { 1, 2, 3, 4, 5, 6, ……. }.
Now as we know that these, r consecutive numbers can start from any number.
So, now as we know that each binomial coefficient gives us an integer value.
So, if n will be any integer that is one smaller than the smallest of r consecutive numbers.
Then, we know that ${}^{n + r}{C_r}$ must be equal to an integer value.
So, let ${}^{n + r}{C_r}$ = m
Now according to binomial expansion, we can write
${}^{n + r}{C_r}$ = $\dfrac{{\left( {{\text{n + r}}} \right)!}}{{\left( {{\text{n + r - r}}} \right)!\left( {\text{r}} \right)!}}$ = $\dfrac{{\left( {{\text{n + r}}} \right)!}}{{\left( {\text{n}} \right)!\left( {\text{r}} \right)!}}$ = m (1)
Now as we know that if x is any natural number then x! will be written as 123*……(x – 2)(x – 1)*(x ).
Now equation 2 can be written as,
$\dfrac{{\left( {{\text{n + r}}} \right)!}}{{\left( {\text{n}} \right)!\left( {\text{r}} \right)!}}$ = $\dfrac{{1 \times 2 \times \times 3 \times 4 \times ....... \times ({\text{n - 1}}) \times ({\text{n}}) \times ({\text{n + 1}}) \times ({\text{n + 2}}) \times ...... \times ({\text{n + r - 2}}) \times ({\text{n + r - 1}}) \times ({\text{n + r}})}}{{\left( {1 \times 2 \times \times 3 \times 4 \times ....... \times ({\text{n}} - 1) \times ({\text{n}})} \right) \times \left( {\text{r}} \right)!}}$ = m
Now as we can see that in the above equation n! is common in numerator and denominator both. So, above equation can also be written as,
$\dfrac{{({\text{n + 1}}) \times ({\text{n + 2}}) \times ({\text{n + 3}}) \times ...... \times ({\text{n + r - 2}}) \times ({\text{n + r - 1}}) \times ({\text{n + r}})}}{{\left( {\text{r}} \right)!}}$ = m (2)
Now as we know that n is one smaller than the smallest of n consecutive numbers.
So, $({\text{n + 1}}) \times ({\text{n + 2}}) \times ({\text{n + 3}}) \times ...... \times ({\text{n + r - 2}}) \times ({\text{n + r - 1}}) \times ({\text{n + r}})$ can also be written as the product of r consecutive integers.
So, now the equation 2 can also be written as,
$\dfrac{{{\text{Product of r consecutive integers}}}}{{\left( {\text{r}} \right)!}}$ = m
As, m is the integer. So, it is true that the product of r consecutive integers is divisible by r!.
Hence, the given statement is true.

Note: Whenever we come up with this type of problem then to find whether the given statement is true or false. We assume an integer m which is equal to the combination ${}^{n + r}{C_r}$. Then we expand the combination and remove n! from numerator and denominator. After that we will get the required result. This will be the easiest and efficient way to find whether the statement is true or false.