# How many sub-committees consisting of five members can be formed from a committee consisting of $ 8 $ gentlemen and $ 6 $ ladies so as to include at least $ 2 $ gentlemen and $ 2 $ ladies?

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**Hint**: Since in this we have to find the combination of five members in which there should be at least $ 2 $ gentleman and $ 2 $ ladies. For this, we will use the concept of combination. So the combination would be like $ 2 $ gentlemen and $ 5 $ ladies or it can be $ 3 $ gentleman and $ 2 $ ladies. And on solving this combination, we will get the answer.

Formula used:

Combination of $ n $ items taken $ r $ at a time

$ c\left( {n,r} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $

Here,

$ n $ , will be the elements of the set

$ r $ , will be the number of elements selected from the set

**:**

__Complete step-by-step answer__Since it is given that there will be five members that can be formed from a committee consisting of $ 8 $ gentlemen and $ 6 $ ladies and also there should be only inclusion of at least $ 2 $ gentleman and $ 2 $ ladies.

So for this, the pair would be like following, $ 2 $ gentleman and $ 3 $ ladies or it can be $ 3 $ gentleman and $ 2 $ ladies.

And mathematically this pair of combination can be written as

$ \Rightarrow \left[ {{}^8{C_2} \times {}^6{C_3}} \right] + \left[ {{}^8{C_3} \times {}^6{C_2}} \right] $

Now by using the combination formula, on substituting the values in the formula we get

\[ \Rightarrow \left[ {\dfrac{{8!}}{{2!\left( {8 - 2} \right)!}} \times \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}}} \right] + \left[ {\dfrac{{8!}}{{3!\left( {8 - 3} \right)!}} \times \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}} \right]\]

Now on solving the small braces first we get

\[ \Rightarrow \left[ {\dfrac{{8!}}{{2!\left( 6 \right)!}} \times \dfrac{{6!}}{{3!\left( 3 \right)!}}} \right] + \left[ {\dfrac{{8!}}{{3!\left( 5 \right)!}} \times \dfrac{{6!}}{{2!\left( 4 \right)!}}} \right]\]

On expanding the above equation for solving, we get

\[ \Rightarrow \left[ {\dfrac{{8 \times 7 \times 6!}}{{2!\left( 6 \right)!}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3 \times 2\left( 3 \right)!}}} \right] + \left[ {\dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times 1\left( 5 \right)!}} \times \dfrac{{6 \times 5 \times 4!}}{{2!\left( 4 \right)!}}} \right]\]

On canceling the like terms and reducing the fractions into the simplest form possible, we get

$ \Rightarrow \left[ {28 \times 20} \right] + \left[ {56 \times 15} \right] $

On solving the multiplication of the above line, we get it as

$ \Rightarrow \left[ {560} \right] + \left[ {840} \right] $

And on adding it, we get

$ \Rightarrow 1400 $

Hence, $ 1400 $ sub-committees consisting of five members can be formed.

**So, the correct answer is “1400”.**

**Note**: For solving this type of problem we need to read the problem properly and then act accordingly. Also, we should know that in combination the order does not matter. So while solving it we should not consider the order. So by using the simple formula we solve such questions.