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A permutation problem: In how many different ways can 2 people be selected from a group of 20 people when the order of selection is important? Complete solution with step-by-step calculations and formula explanation.

Problem

In how many different ways can 2 people be selected from a group of 20 people when the order of selection is important?

Key Concepts

  • Permutation: An arrangement where order matters.
  • Permutation Formula: When selecting rr objects from nn objects, the number of permutations is P(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}
  • Factorial Notation: n!n! represents n×(n−1)×(n−2)×...×2×1n \times (n-1) \times (n-2) \times ... \times 2 \times 1

Solution

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Since the order of selection is important (for example, choosing person A first and person B second is different from choosing person B first and person A second), we need to use the permutation formula:

Pnr=n!(n−r)!P_n^r = \frac{n!}{(n-r)!}

In this problem:

  • n=20n = 20 (total number of people)
  • r=2r = 2 (number of people to select)

Substituting these values into the formula:

P202=20!(20−2)!=20!18!P_{20}^2 = \frac{20!}{(20-2)!} = \frac{20!}{18!}

We can simplify this expression:

20!18!=20×19×18!18!=20×19=380\frac{20!}{18!} = \frac{20 \times 19 \times 18!}{18!} = 20 \times 19 = 380

Therefore, there are 380 different ways to select 2 people from a group of 20 people when the order matters.

Applications

This type of problem appears in various scenarios:

  • Selecting first and second place winners in a competition
  • Assigning president and vice-president roles from a committee
  • Choosing primary and backup representatives from a group
  • Planning sequential interviews from a pool of candidates
  • Permutation with repetition allowed: nrn^r
  • Circular permutation: (n−1)!(n-1)!
  • Combination formula (when order doesn’t matter): C(n,r)=(nr)=n!r!(n−r)!C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}
# Statistics # Mathematics # Probability # Permutations # Combinatorics

Frequently Asked Questions

What is the difference between permutation and combination? ▼

In a permutation, the order of selection matters (AB is different from BA), while in a combination, the order doesn't matter (AB is the same as BA). Permutations are used when arranging objects in order, while combinations are used when selecting objects without regard to order.

How do you calculate the number of permutations? ▼

The formula for calculating permutations of $r$ objects selected from $n$ objects is $P(n,r) = \frac{n!}{(n-r)!}$. This counts all possible arrangements where order matters.

What is the value of P(20,2)? ▼

P(20,2) = $\frac{20!}{(20-2)!}$ = $\frac{20!}{18!}$ = $20 \times 19$ = 380. This means there are 380 different ways to select and arrange 2 people from a group of 20 people.