Question

If $n{C_r} = 10$, $n{C_{r + 1}} = 45$ then, $r$ equals to
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: In this problem, we have given the value of $n$ number of different items and we have to choose $r$ number of items from it. And also we have given the value of $n$ number of different items and we have to choose $r + 1$ number of items from it. Here our aim is to find the value of $r$. To solve this problem we are going to use the combination formula and by using this formula we find the value of $r$.

Formula used: $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where,
$n$= the number of items and
$r$=how many items are taken at a time.
$n{C_{r + 1}} = \dfrac{{n!}}{{(r + 1)!(n - r - 1)!}}$
Where,
$n$= the number of items and
$r + 1$=how many items are taken at a time.

Complete step-by-step solution:
Here it is given in the question that, $n{C_r} = 10$, $n{C_{r + 1}} = 45$
That is, $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} = 10$
And $n{C_{r + 1}} = \dfrac{{n!}}{{(r + 1)!(n - r - 1)!}} = 45$
Let us consider that,
$\dfrac{{n{C_{r + 1}}}}{{n{C_r}}} = \dfrac{{\dfrac{{n!}}{{\left( {r + 1} \right)!(n - r - 1)!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}}$
Hence we get,
$ \Rightarrow \dfrac{{\dfrac{{n!}}{{\left( {r + 1} \right)!(n - r - 1)!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}} = \dfrac{{45}}{{10}}$
Solving this we get,
$ \Rightarrow \dfrac{{n - r}}{{r + 1}} = \dfrac{{45}}{{10}} - - - - - \left( * \right)$
Solving the numerical fraction in the right hand side, we get
$ \Rightarrow \dfrac{{n - r}}{{r + 1}} = \dfrac{9}{2}$
Cross multiply the above equation we get,
$2(n - r) = 9(r + 1)$
$2n - 2r = 9r + 9$
Keep the r terms in one side, we get
$2n = 11r + 9 - - - - - (1)$
Now let’s give the values for $r$
If $r = 2{\text{ or }}4$ then equation 1 becomes,
$r = 2 \Rightarrow n = \dfrac{{31}}{2}$ is not an integer.
$r = 4 \Rightarrow n = \dfrac{{53}}{2}$ is not an integer.
If $r = 3$ then $n = \dfrac{{42}}{2} = 21$
So in the case $r = 3$ we get $n = 21$, and $n{C_r} = 21{C_3} \ne 10$
So $r = 3$ is not a correct answer.
If $r = 1 \Rightarrow n = \dfrac{{20}}{2} = 10$ which is an integer
Then in the case when $r = 1\& n = 10$ the combination formula $n{C_r} = 10{C_1} = 10$ and also $n{C_{r + 1}} = 10{C_2} = 45$
Therefore, $r = 1$ and $n = 10$ is the correct answer.

Hence, the answer is option (A)

Note: We have to remember that, in mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter (unlike permutations).
We can write $\left( {r + 1} \right)!$cas $r!(r + 1)$also we can write $\left( {n - r} \right)!$ as $(n - r)(n - r - 1)!$. This is the main thing we used in our problem to bring up the required solution. Equation (*) is claimed by using these formulas and cancelling all the same terms in the numerator and in the denominator. And also we substituted the value of r and n in the $n{C_r}$ and $n{C_{r + 1}}$ formulas to get the given required solution.