Combinatorics
PYP_JEE_ADV_2024_P1
Grade None
Question:
A group of 9 students, $s_1,s_2,\ldots,s_9$, is to be divided to form three teams $X$, $Y$, and $Z$ of sizes 2, 3, and 4, respectively. Suppose that $s_1$ cannot be selected for the team $X$, and $s_2$ cannot be selected for the team $Y$. Then the number of ways to form such teams is ___.
Step-by-Step Solution
Key Concept: Inclusion-exclusion: subtract bad cases then add back over-subtracted intersection
Total (unrestricted): $\binom{9}{2}\binom{7}{3}=36\times35=1260$.
Let $A$=event $s_1\in X$, $B$=event $s_2\in Y$.
$|A|$: fix $s_1$ in $X$, choose 1 more from 8: $\binom{8}{1}\binom{7}{3}=8\times35=280$.
$|B|$: fix $s_2$ in $Y$, choose 2 for $X$ from 8 (excluding $s_2$): $\binom{8}{2}\binom{6}{2}=28\times15=420$.
Wait: if $s_2$ in $Y$, choose 2 for $X$ from remaining $\{s_1,s_3,\ldots,s_9\}$ (8 students), then 2 more for $Y$ from remaining 7: $\binom{8}{2}\binom{7}{2}=28\times21=588$.
$|A\cap B|$: $s_1\in X$, $s_2\in Y$. Choose 1 more for $X$ from $\{s_3,\ldots,s_9\}$ (7 students): $\binom{7}{1}=7$. Choose 2 more for $Y$ from remaining 6: $\binom{6}{2}=15$. Total: $7\times15=105$.
Answer $=1260-280-588+105=497$.
Correct Answer: 497