Question

4 identical hats are distributed among 4 boys such that any boy can get any number of hats. What is the probability that no boys gets more than 2 hats.

• A. 18/35
• B. 2/7
• C. 13/35
• D. 16/35

Answer 1

The correct option is C

13/35

Total ways to distribute the bags is = Number of solutions of the equation

$a+b+c+d=4$

or, ${}^{(4+4-1)}{C}_{(4-1)}{=}^7{C}_3=35$

When maximum 2 hats can be distributed

Following cases is possible: $2,1,1,0{=}^4{C}_{1}X{}^3{C}_1=12$

or, $1,1,1,1=1$ way.

$P\left(E\right)=\frac{13}{35}$