AMC 10 Daily Practice - Counting

Complete problem set with solutions and individual problem pages

Problem 1 Easy

There are 8 different pens to be distributed to the top three students, with each student receiving one pen. There are            different ways to distribute the pens.

Answer:336

_8P_3=8\times7\times6=336.

Link Problem
Problem 2 Easy

Using numbers 1, 2, 3, 4, 5 and, 6 to form a 6 digit number without repeating digits. 1 and 2 must be adjacent, and 3 and 4 must be non-adjacent. In how many ways can this happen?

  • A.

    72

  • B.

    144

  • C.

    216

  • D.

    288

  • E.

    312

Answer:B

Binding 1 and 2: _2P_2 ways to arrange them.

Arrange 1,2, 5, and 6: _3P_3 ways to arrange them.

Use star and bar method to place 3 and 4: {_4P_2} ways to fill in 3 and 4.

In total, there are _2P_2 \times _3P_3 \times _4P_2 = 144 ways.

Link Problem
Problem 3 Easy

Ang, Ben, and Jasmin each have 5 blocks, colored red,blue, yellow, white, and green; andthere are 5 empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives 3 blocks all of the same color is \dfrac{m}{n},where m and n are relatively prime positive integers. What is m+n? (2021 AMC Spring 10B Problems, Question #22)

  • A.

    47

  • B.

    94

  • C.

    227

  • D.

    471

  • E.

    542

Answer:E

Let our denominator be (5!)^3, so we consider all possible distributions.

We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.

When we have at 1 box with all 3 balls the same color in that box, there are _{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 ways for the distributions to occur (_{5} C _{1} for selecting one of the five boxes for a uniform color, _{5} P _{1} for choosing the color for that box, 4! for each of the three people to place their remaining items).

However, we overcounted those distributions where two boxes had uniform color, and there are _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 ways for the distributions to occur (_{5} C _{2} for selecting two of the five boxes for a uniform color, _{5} P _{2} for choosing the color for those boxes, 3! for each of the three people to place their remaining items).

Again, we need to re-add back in the distributions with three boxes of uniform color\cdots and so on so forth.

Our success by PIE is

_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 - _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 + _{5} C _{3} \cdot _{5} P _{3} \cdot (2!)^3 - _{5} C _{4} \cdot _{5} P _{4} \cdot (1!)^3 + _{5} C _{5} \cdot _{5} P _{5} \cdot (0!)^3 = 120 \cdot 2556.\frac{120 \cdot 2556}{120^3}=\frac{71}{400},

yielding an answer of \boxed{\textbf{(D) }471}.

Link Problem
Table of Contents
  • 1
  • 2
  • 3