Question

All possible two factor’s products are formed from numbers \[1,{\rm{ }}2,{\rm{ }}3,{\rm{ }}4,{\rm{ }}....,{\rm{ }}200\]. Find the number of factors out of the total obtained factors which are multiples of 5.
A. \[5040\]
B. \[7180\]
C. \[8150\]
D. None of these

Answer 1

Hint: First, calculate the total number of two factor products. Then, find the number of numbers which are not multiples of 5 from 1 to \[200\]. After that, calculate the total number of two factor products which are not multiple of 5. In the end, subtract the total number of two factor products which are not multiple of 5 from the total number of two factor products and get the required answer.



Formula Used:The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Factorial: \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]



Complete step by step solution:The given numbers are \[1,{\rm{ }}2,{\rm{ }}3,{\rm{ }}4,{\rm{ }}....,{\rm{ }}200\].
The total number of two factor products are: \[{}^{200}{C_2}\] \[.....\left( 1 \right)\]

Now find the number of numbers which are not multiples of 5 from 1 to \[200\].
We know that, \[5,{\rm{ 10}},{\rm{ 15}},{\rm{ 20}},{\rm{ }}....,{\rm{ }}200\] are the \[40\] multiples of 5.
So, the numbers from 1 to \[200\] which are not multiples of 5: \[160\]
Therefore, the number of two-factor products which are not multiples 5: \[{}^{160}{C_2}\] \[.....\left( 2 \right)\]

To calculate the number of factors out of the total obtained factors which are multiples of 5, subtract equation \[\left( 2 \right)\] from the equation \[\left( 1 \right)\].
We get,
\[{}^{200}{C_2} - {}^{160}{C_2} = \dfrac{{200!}}{{2!\left( {200 - 2} \right)!}} - \dfrac{{160!}}{{2!\left( {160 - 2} \right)!}}\]
\[ \Rightarrow {}^{200}{C_2} - {}^{160}{C_2} = \dfrac{{200!}}{{2!198!}} - \dfrac{{160!}}{{2!158!}}\]
Simplify the right-hand side by applying the formula \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]
\[ \Rightarrow {}^{200}{C_2} - {}^{160}{C_2} = \dfrac{{200 \times 199 \times 198!}}{{2!198!}} - \dfrac{{160 \times 159 \times 158!}}{{2!158!}}\]
\[ \Rightarrow {}^{200}{C_2} - {}^{160}{C_2} = \dfrac{{200 \times 199}}{2} - \dfrac{{160 \times 159}}{2}\]
\[ \Rightarrow {}^{200}{C_2} - {}^{160}{C_2} = 19900 - 12720\]
\[ \Rightarrow {}^{200}{C_2} - {}^{160}{C_2} = 7180\]
Thus, the number of factors which are multiples of 5 is \[7180\].



Option ‘B’ is correct



Note: Students often get confused and consider the combination formula as \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], which is an incorrect formula. Because of this, they get the wrong solution.