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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

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.

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