AMC10 2010 B

AMC10 2010 B · Q22

AMC10 2010 B · Q22. It mainly tests Basic counting (rules of product/sum), Inclusion–exclusion (basic).

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

有7块不同的糖果要分给三个袋子。红袋和蓝袋各至少得一块糖果；白袋可以为空。可能的分法有多少种？

(A) 1930 1930

(B) 1931 1931

(D) 1933 1933

(E) 1934 1934

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): If there were no restrictions on the number of candies per bag, then each piece of candy could be distributed in 3 ways. In this case there would be $3^7$ ways to distribute the candy. However, this counts the cases where the red bag or blue bag is empty. If the red bag remained empty then the candy could be distributed in $2^7$ ways. The same is true for the blue bag. Both totals include the case in which all the candy is put into the white bag. Hence there are $2^7 + 2^7 - 1$ ways to distribute the candy such that either the red or blue bag is empty. The number of ways to distribute the candy, subject to the given conditions, is $3^7 - (2^7 + 2^7 - 1) = 1932$.

答案（C）：如果对每个袋子里糖果的数量没有限制，那么每块糖果都有 3 种分配方式。在这种情况下，分配糖果的方法共有 $3^7$ 种。然而，这样会把红袋或蓝袋为空的情况也计算进去。如果红袋保持为空，则糖果可以用 $2^7$ 种方式分配。蓝袋同理。这两个总数都包含“所有糖果都放入白袋”的情况。因此，使得红袋或蓝袋至少一个为空的分配方式共有 $2^7 + 2^7 - 1$ 种。在给定条件下，分配糖果的方法数为 $3^7 - (2^7 + 2^7 - 1) = 1932$。

Topics