Question

The probability that the birthdays of 6 boys will fall exactly in 3 calender months is

• A. $\frac{{}^{12}{C}_3(3^6-3)}{{12}^6}$
• B. $\frac{{}^{12}{C}_3\left(3^6\right)}{{12}^6}$
• C. $\frac{{}^{12}{C}_3\left(192\right)}{{12}^6}$
• D. $\frac{{}^{12}{C}_3\left(540\right)}{{12}^6}$

Answer 1

Answer: (D)

$\frac{{}^{12}{C}_3\left(540\right)}{{12}^6}$

Total cases here is : ${12}^6$ as each boy has 12 choices and 6 boys.
Now if we want them to have their birthdays in exactly $3$ months, we need to choose the $3$ months first. this can be done in ${}^{12}{C}_3$ ways.
Do this and then each boy has 3 choices, month A, month B or month C. $⟹3^6$
but we also need to subtract the cases in which all of them have their birthdays in one month or 2 months exactly.

Number of cases for exactly two months = ${}^3{C}_2(2^6-2)$

Number of cases for exactly one month = ${}^3{C}_1$

So we subtract these cases $⟹3^6{-}^3{C}_2(2^6-2){-}^3{C}_1=540$
Hence the given answer.