Question

There are three cycles to distributed among five children. If no child gets more than one cycle, then this can be done in how many ways?

• A. 10
• B. 30
• C. 60
• D. None of the above

Answer 1

Answer:

C. 60

Understanding the Problem: Permutations

• This problem involves selecting a specific number of items (cycles) and arranging them among a smaller group (children), where the order of selection matters, and no repetition is allowed. This is a classic permutation scenario.

• The condition "no child gets more than one cycle" is crucial as it signifies that once a child receives a cycle, they are out of the pool for receiving another, which is a key characteristic of permutations without repetition.

Key Concepts & Formula:

• When we need to select r items from a set of n distinct items and arrange them, the number of ways is given by the permutation formula:
  P(n, r) = n! / (n - r)!

• In this problem:

  • n = Total number of children available = 5

  • r = Number of cycles to be distributed (which also represents the number of children who will receive a cycle) = 3

Step-by-Step Calculation:

1. Identify 'n' and 'r' from the problem statement: n=5, r=3.

2. Apply the permutation formula: P(5, 3) = 5! / (5 - 3)!

3. Simplify the expression: P(5, 3) = 5! / 2!

4. Calculate the factorials:

   • 5! = 5 × 4 × 3 × 2 × 1 = 120

   • 2! = 2 × 1 = 2

5. Perform the division: P(5, 3) = 120 / 2 = 60.

6. Therefore, there are 60 different ways to distribute the three cycles among five children such that no child gets more than one cycle.