Question

$10$ different books and $2$ different pens are given to $3$ boys so that each gets equal number of things. The probability that the same boy does not receive both the pens is

• A. $\frac{5}{11}$
• B. $\frac{7}{11}$
• C. $\frac{2}{3}$
• D. $\frac{8}{11}$

Answer 1

The correct option is D $\frac{8}{11}$

Probability of same boy doesn't receive both pens $=1-$ probability of any boy receive both pens.
Since all the students recieves and number of article. So each one get $\frac{(10+2)}{3}=4$
Number of ways in which $10$ books and $2$ pens is distributed to $3$ students ${}^{12}{C}_4{.}^8{C}_4{.}^4{C}_4$
Now find the number of ways such that any boy receive the both pen.
Select one boy from $3$ and give both pen to him in ${3e}_1\times 1$ ways.
Rest two article can be given to this boy in ${}^{10}{C}_2$ ways.
Other $2$ boys will get article in ${}^8{C}_4{\times }^4{C}_4$ ways.
So number of ways such that any boy receive the both pen ${}^3{C}_1\times 1{\times }^{10}{C_2\times }^8{C}_2{\times }^4{C}_4$
So required probability $=\frac{{1.}^3{C}_1\times 1{\times }^{10}{C_2\times }^8{C_4\times }^4{C}_4}{{}^{12}{C_4\times }^8{C_4\times }^4{C}_4}=\frac{8}{11}$