Find the value r from the following combination ${}^8{C_r} - {}^7{C_3} = {}^7{C_2}$ .
Answer
640.8k+ views
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
${}^8{C_r} = {}^7{C_2} + {}^7{C_3}$…………….. (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
${}^8{C_r} = {}^7{C_2} + {}^7{C_3}$…………….. (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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

