MathFun4Kids

Solution Let’s assume function \(f_1(n)\) is number of ways using pennies to make a change of \(n\) cents. Obviously, we have $$f_1(n) = \begin{cases} 0 & \text { if n < 0 }\\ 1 & \text { if n = 0 }\\ f_1(n-1) = \cdots = f_1(0) = 1 & \text { if n > 0 } \end{cases} $$ Lets assume function \(f_2(n)\) is number of ways using pennies and/or nickels to make a change of \(n\) cents. $$f_2(n) = \begin{cases} 0 & \text { if n < 0 }\\ 1 & \text { if n = 0 }\\ f_1(n)+f_2(n-5) = f_2(n-5) +1 & \text { if n > 0 } \end{cases} $$ Lets assume \(f_3(n)\) is number of ways using pennies, nickels and/or dimes to make a change of \(n\) cents. $$f_3(n) = \begin{cases} 0 & \text { if n < 0 }\\ 1 & \text { if n = 0 }\\ f_2(n)+f_3(n-10) & \text { if n > 0 } \end{cases} $$ Lets assume \(f_4(n)\) is number of ways using pennies, nickels, dimes and/or quarters to make a change of \(n\) cents. $$f_4(n) = \begin{cases} 0 & \text { if n < 0 }\\ 1 & \text { if n = 0 }\\ f_3(n)+f_4(n-25) & \text { if n > 0 } \end{cases} $$ Obviously, the answer for making a changes of 63 cents is the same as to make 60 cents, because 3 pennies must be used to make a change of 3 cents. So we consider \(n\) that are multiples of 5, as shown in the following table: $$ \begin{array}{c|c|c|c|c} \ n\ & f_1(n) & f_2(n) & f_3(n) & f_4(n) \\ \hline 0 & 1 & 1 & 1 & 1\\ 5 & 1 & 2 & 2 & 2\\ 10 & 1 & 3 & 4 & 4\\ 15 & 1 & 4 & 6 & 6\\ 20 & 1 & 5 & 9 & 9\\ 25 & 1 & 6 & 12 & 13\\ 30 & 1 & 7 & 16 & 18\\ 35 & 1 & 8 & 20 & 24\\ 40 & 1 & 9 & 25 & 31\\ 45 & 1 & 10 & 30 & 39\\ 50 & 1 & 11 & 36 & 49\\ 55 & 1 & 12 & 42 & 60\\ 60 & 1 & 13 & 49 & 73\\ %65 & 1 & 14 & 56 & 87\\ \end{array} $$ Therefore, the answer to the question is: $$ f_4(63) = f_4(60) = 73 $$ \(\blacksquare\)

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: $$\sum_{n=0}^{3}{C(12, n)\cdot C(12-n,3-n)} $$ \(=C(12,0)\cdot C(12,3)+C(12,1)\cdot C(11,2)+C(12,2)\cdot C(10,1)+C(12,3)\cdot C(9,0)\)
\(=220+660+660+220=1760\)
\(\blacksquare\)