UPSEE Solved Model Paper 1

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 = $\frac{3!}{2!} \cdot \cdot \,{}^{6}C_4 \cdot \,{}^{2}C_1 \cdot \,{}^{1}C_1$ = 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 = $\frac{3!}{3!} \,{}^{6}C_2 \cdot \,{}^{4}C_2 \cdot \,{}^{2}C_2$ = 90 ways. The number of ways so that the birthdays of 6 people falls in exactly 3 calendar months = (90+360+ 90)- 220 = 118800