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Hint: In this question apply the property of combination, later on compare the values, so use these concepts to get the solution of the question.

Given equation is

${}^8{C_r} - {}^7{C_3} = {}^7{C_2}$

Above equation is also written as

${}^8{C_r} = {}^7{C_2} + {}^7{C_3}$â€¦â€¦â€¦â€¦â€¦.. (1)

Now we all know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , so use this property in above equation we have,

${}^8{C_r} = \dfrac{{8!}}{{r!\left( {8 - r} \right)!}}$

${}^7{C_2} = \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} = \dfrac{{7!}}{{2! \times 5!}}$

${}^7{C_3} = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}} = \dfrac{{7!}}{{3! \times 4!}}$

Therefore from equation (1) we have,

$\dfrac{{8!}}{{r!\left( {8 - r} \right)!}} = \dfrac{{7!}}{{2! \times 5!}} + \dfrac{{7!}}{{3! \times 4!}}$

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

Now we all know that $\left( {8 \times 7! = 8!} \right),\left( {3 \times 2! = 3!} \right),\left( {5 \times 4! = 5!} \right)$

$

\Rightarrow \dfrac{{8!}}{{r!\left( {8 - r} \right)!}} = \dfrac{{8!}}{{3! \times 5!}} \\

\Rightarrow \dfrac{{8!}}{{r!\left( {8 - r} \right)!}} = \dfrac{{8!}}{{3! \times \left( {8 - 3} \right)!}} \\

$

Now compare the denominator part of L.H.S and R.H.S as numerator is same we have,

$r = 3$.

So, this is the required value of r.

Note: In such types of questions the key concept we have to remember is that always recall the property of combination which is stated above, then using this property simplify the equation and then compare the denominator part of L.H.S and R.H.S as numerator is same, we will get the required value of r.

Given equation is

${}^8{C_r} - {}^7{C_3} = {}^7{C_2}$

Above equation is also written as

${}^8{C_r} = {}^7{C_2} + {}^7{C_3}$â€¦â€¦â€¦â€¦â€¦.. (1)

Now we all know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , so use this property in above equation we have,

${}^8{C_r} = \dfrac{{8!}}{{r!\left( {8 - r} \right)!}}$

${}^7{C_2} = \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} = \dfrac{{7!}}{{2! \times 5!}}$

${}^7{C_3} = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}} = \dfrac{{7!}}{{3! \times 4!}}$

Therefore from equation (1) we have,

$\dfrac{{8!}}{{r!\left( {8 - r} \right)!}} = \dfrac{{7!}}{{2! \times 5!}} + \dfrac{{7!}}{{3! \times 4!}}$

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

Now we all know that $\left( {8 \times 7! = 8!} \right),\left( {3 \times 2! = 3!} \right),\left( {5 \times 4! = 5!} \right)$

$

\Rightarrow \dfrac{{8!}}{{r!\left( {8 - r} \right)!}} = \dfrac{{8!}}{{3! \times 5!}} \\

\Rightarrow \dfrac{{8!}}{{r!\left( {8 - r} \right)!}} = \dfrac{{8!}}{{3! \times \left( {8 - 3} \right)!}} \\

$

Now compare the denominator part of L.H.S and R.H.S as numerator is same we have,

$r = 3$.

So, this is the required value of r.

Note: In such types of questions the key concept we have to remember is that always recall the property of combination which is stated above, then using this property simplify the equation and then compare the denominator part of L.H.S and R.H.S as numerator is same, we will get the required value of r.

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