AMC12 2021 A

AMC12 2021 A · Q15

AMC12 2021 A · Q15. It mainly tests Combinations, Remainders & modular arithmetic.

A choir director must select a group of singers from among his $6$ tenors and $8$ basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of $4$, and the group must have at least one singer. Let $N$ be the number of different groups that could be selected. What is the remainder when $N$ is divided by $100$?

合唱团指挥必须从$6$名男高音和$8$名男低音中选出一组歌手。唯一要求是男高音和男低音的数量之差必须是$4$的倍数，且组中至少有一名歌手。设$N$是可以选出的不同组的数量。$N$除以$100$的余数是多少？

(A) 47 47

(B) 48 48

(D) 95 95

(E) 96\qquad 96\qquad

Answer

Correct choice: (D)

正确答案：（D）

Solution

Suppose that $t$ tenors and $b$ basses are selected. The requirements are $t\equiv b\pmod{4}$ and $(t,b)\neq(0,0).$ It follows that $b'=8-b$ basses are not selected. Since the ordered pairs $(t,b)$ and the ordered pairs $(t,b')$ have one-to-one correspondence, we consider the ordered pairs $(t,b')$ instead. The requirements become $t\equiv8-b'\pmod{4}$ and $(t,8-b')\neq(0,0),$ which simplify to $t+b'\equiv0\pmod{4}$ and $(t,b')\neq(0,8),$ respectively. As $t+b'\in\{0,4,8,12\},$ the total number of such groups is \begin{align*} N&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{8}-1\right]+\binom{14}{12} \\ &=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{6}-1\right]+\binom{14}{2} \\ &=1+1001+[3003-1]+91 \\ &=4095, \end{align*} from which $N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.$

设选$t$名男高音和$b$名男低音。要求$t\equiv b\pmod{4}$且$(t,b)\neq(0,0)$。因此，未选$b'=8-b$名男低音。由于有序对$(t,b)$和$(t,b')$一一对应，我们考虑有序对$(t,b')$。要求变为$t\equiv8-b'\pmod{4}$且$(t,8-b')\neq(0,0)$，简化为$t+b'\equiv0\pmod{4}$和$(t,b')\neq(0,8)$。由于$t+b'\in\{0,4,8,12\}$，此类组的总数为 \begin{align*} N&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{8}-1\right]+\binom{14}{12} \\ &=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{6}-1\right]+\binom{14}{2} \\ &=1+1001+[3003-1]+91 \\ &=4095, \end{align*} 因此$N\equiv\boxed{\textbf{(D) } 95}\pmod{100}$。

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