Question

The sum $\sum\limits_{i=0}^{m}{C_{i}^{10}}\times C_{m-i}^{20}$(where$C_{q}^{p}=0$, if$p < q$) is maximum when $m$ is?
A. $5$
B. $15$
C. $10$
D. $20$

Answer 1

To solve these types of questions, firstly we have to understand the formula of combination used for a large number of groups of elements. After that we have to substitute values according to our given question and then we have to identify the condition at which we can get our maximum value and in this way we will get our required value.

Complete step by step answer:
Combination is a way of selecting items from a collection where the order of selection does not matter. The combination can be defined as “an arrangement of objects where the order of selected objects does not matter”.
For example: If we want to buy a milkshake and we are allowed to combine any two flavors from Apple, Banana and Cherry. So we are supposed to make a combination out of these possible flavors. We have total combinations as: Apple and banana, Apple and Cherry, Banana and Cherry. These are the only possible combinations from the given flavors and as we can see in this order of selected flavors do not matter.
As in smaller cases, it is possible for us to count the number of combinations, but when the cases have a large number of groups of elements, then we use a formula to determine all the possible selections. The formula is represented as:
$^{\mathbf{n}}{{\mathbf{C}}_{\mathbf{k}~}}=\text{ }\left[ \left( \mathbf{n} \right)\left( \mathbf{n}-\mathbf{1} \right)\left( \mathbf{n}-\mathbf{2} \right)\ldots .\left( \mathbf{n}-\mathbf{k}+\mathbf{1} \right) \right]/\left[ \left( \mathbf{k}-\mathbf{1} \right)\left( \mathbf{k}-\mathbf{2} \right)\ldots \ldots .\left( \mathbf{1} \right) \right]$
Now we have given in the question that:
$\sum\limits_{i=0}^{m}{C_{i}^{10}}\times C_{m-i}^{20}$
So now we will try to expand this expression by substituting values of $i$from $0$ to$m$.
After expanding the given expression we get as:
$sum{{=}^{10}}{{\mathbf{C}}_{0~}}{{.}^{20}}{{\mathbf{C}}_{m~}}{{+}^{10}}{{\mathbf{C}}_{1~}}{{.}^{20}}{{\mathbf{C}}_{m-1~}}+.............{{+}^{10}}{{\mathbf{C}}_{m~}}{{.}^{20}}{{\mathbf{C}}_{0~}}$
Now in such cases we have to consider two binomial expansions ${{(1+x)}^{p}}$ and ${{(1+x)}^{q}}$.
Here$p=10,q=20$. Now we will multiply them and collect the required exponent of $x$ to which we obtain the given expression as:
$sum={{(1+x)}^{10}}.{{(1+x)}^{20}}.{{x}^{m}}$
$sum{{=}^{30}}{{\mathbf{C}}_{m~}}$
There is a condition, that $^{n}{{\mathbf{C}}_{r~}}$is maximum only when $r=\dfrac{n}{2}$.
So for $^{30}{{\mathbf{C}}_{m~}}$ to be maximum, it is required that $m=\dfrac{30}{2}$
Therefore, $m=\dfrac{30}{2}=15$ is our required maximum value.

So, the correct answer is “Option B”.

Note: The difference between the permutation and combination is only of the order of the objects. In permutation the order of objects is very important, i.e. the objects must be in the proper order. In permutation there are only ordered elements whereas in the combination it can have unordered sets.