Question

The number of ways so that the birthdays of 6 people falls in exactly 3 calendar months is

• A. 118800
• B. 36960
• C. 158400
• D. 47520

Answer 1

The correct option is A

118800

No. of ways to choose the 3 months $={{}^{12}C}_3=220.$

Now, we have 6 people and 3 months for the birthdays to fall in.

Since none of these three months can have no birthday in it, the 3 possible ways of distributing the 6 people in 3 months would be $(4,1,1),(3,2,1),(2,2,2)$

$(4,1,1)$

No. of ways of rotating the distribution of number of birthdays in these three months $=\frac{3!}{2!}$

No. of ways of rotating the distribution of the people $={{}^{6}C}_4\cdot {{}^{2}C}_1\cdot {{}^{1}C}_1$

Therefore, total number of ways in $(4,1,1)$ case $=\frac{3!}{2!}\cdot {{}^{6}C}_4\cdot {{}^{2}C}_1\cdot {{}^{1}C}_1=90$ ways.

$(3,2,1)$

No. of ways of rotating the distribution of number of birthdays in these three months $=3!$

No. of ways of rotating the distribution of the people $={{}^{6}C}_3\cdot {{}^{3}C}_2\cdot {{}^{1}C}_1$

Therefore, total number of ways in $(3,2,1)$ case $=3!\cdot {{}^{6}C}_3\cdot {{}^{3}C}_2\cdot {{}^{1}C}_1=360$ ways.

$(2,2,2)$

No. of ways of rotating the distribution of number of birthdays in these three months $=\frac{3!}{3!}$

No. of ways of rotating the distribution of the people $={{}^{6}C}_2\cdot {{}^{4}C}_2\cdot {{}^{2}C}_2$

Therefore, total number of ways in $(2,2,2)$ case $=\frac{3!}{3!}\cdot {{}^{6}C}_2\cdot {{}^{4}C}_2\cdot {{}^{2}C}_2=90$ ways.

The number of ways so that the birthdays of 6 people falls in exactly 3 calendar months $=(90+360+90)\cdot 220=118800$