MATHCOUNTS Exercises – 5

Solution 1 The total number of 3-person groups can be formed without restriction is: $$C(24,3)=\frac{24!}{3!\times (24-3)!}=\frac{24\times 23\times 22}{3\times 2 \times 1}=2024$$ The total number of 3-person groups with two persons as a couple is: $$C(12,1)\times C(22,1)=12\times 22=264$$ Therefore, the total number of different teams formed without two people as a couple is: $$2024-264=1760$$ \(\blacksquare\)

Solution 2 Numbering the couples from 1 to 12, the total number of ways to choose 3 couples from 12 is: $$C(12,3)=\frac{12!}{3!\times(12-3)!}=\frac{12\times 11\times 10}{3\times 2\times 1}=220$$ For each couple among 3 chosen couples, either the male or female can be chosen, therefore, the answer to the question is $$220\times (2\times 2\times 2) = 1760$$\(\blacksquare\)

Solution 3 There are 24 ways to choose the first person, 22 ways to choose the second person by excluding the spouse of the first person, and 20 ways to choose the third person, by excluding the spouses of the first two persons, to form a 3-person team. Therefore, the total number of different teams formed without two people as a couple is: $$\frac{24\times 22\times 20}{3!}=\frac{10560}{6}=1760$$\(\blacksquare\)
Solution 4 The number of ways to form a 3-person team with \(n\) males and \(3-n\) females from 12 couples is $$C(12, n)\cdot C(12-n,3-n)$$ Therefore the total ways to form the team is: $$\sum_{n=0}^{3}{C(12, n)\cdot C(12-n,3-n)} $$ \(=C(12,0)\cdot C(12,3)+C(12,1)\cdot C(11,2)+C(12,2)\cdot C(10,1)+C(12,3)\cdot C(9,0)\)
\(=220+660+660+220=1760\)
\(\blacksquare\)