2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 3 Easy

In the expression (            \times            )+(            \times            ) each blank is to be filled in with one of the digits 1, 2, 3, or 4, with each digit being used once. How many different values can be obtained? (2018 AMC 10B Problem, Question#3)

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    6

  • E.

    24

Answer:B

We have \left( \begin{array}{l} {4}\\{2} \end{array}\right) ways to choose the pairs, and we have 2 ways for the values to be switched so \frac{6}{2}= 3.

We have four available numbers (1,2,3,4).Because different permutations do not matter because they are all addtion and multiplication, if we put 1 on the first space, it is obvious there are possible outcomes (2,3,4).

There are 4! ways to arrange the numbers and 2!2!2! overcounts per way due to commutativity.

Therefore, the answer is \dfrac{4!}{2!2!2!}=3.

Related Problems
Problem 1 2018 AMC 10 B
Problem 2 2018 AMC 10 B
Problem 4 2018 AMC 10 B
Problem 5 2018 AMC 10 B
Think Academy Powered by ChatGPT