Mathematics - Permutation & Combination Question with Solution | TestHub

For (P): The letters are M(1), I(4), S(4), P(2). First, arrange the 7 letters other than 'S' (M, I, I, I, I, P, P) in $\frac{7!}{4!2!}=105$ ways. These 7 letters create 8 gaps. To ensure no two 'S's are adjacent, place the 4 'S's in 4 distinct gaps in $C(8,4)=70$ ways. Total permutations = $105\times 70=7350$.

For (Q): We need solutions to $x_1+x_2+x_3=10$ with $1\le x_i\le 5$. Let $y_i=x_i-1$, so $y_1+y_2+y_3=7$ with $0\le y_i\le 4$. Total non-negative solutions are $C(7+3-1,3-1)=C(9,2)=36$. Using inclusion-exclusion, let $A_i$ be the condition $y_i\ge 5$. $|A_1|=C((7-5)+3-1,3-1)=C(4,2)=6$. So, $|A_1|+|A_2|+|A_3|=3\times 6=18$. No solution exists for $y_i\ge 5$ and $y_j\ge 5$ simultaneously. Thus, $36-18=18$.

For (R): Total ways to divide 9 distinct players into 3 teams of 3 is $\frac{C(9,3)C(6,3)C(3,3)}{3!}=\frac{84\times 20\times 1}{6}=280$. If two specific players (A and B) are in the same team, choose 1 more player for their team from the remaining 7 in $C(7,1)=7$ ways. The remaining 6 players form 2 teams of 3 in $\frac{C(6,3)C(3,3)}{2!}=10$ ways. So, A and B are in the same team in $7\times 10=70$ ways. Thus, A and B are in different teams in $280-70=210$ ways.

For (S): Total circular arrangements of 8 people is $(8-1)!=7!=5040$. Using inclusion-exclusion, let $C_i$ be the condition that couple $i$ sits together. $S_1=C(4,1)\times 2!\times 6!=5760$. $S_2=C(4,2)\times (2!{)}^2\times 5!=2880$. $S_3=C(4,3)\times (2!{)}^3\times 4!=768$. $S_4=C(4,4)\times (2!{)}^4\times 3!=96$. The number of arrangements is $7!-S_1+S_2-S_3+S_4=5040-5760+2880-768+96=828$.