##### Permutations

**THEOREM 1** If n is a positive integer and r is an integer with 1 ≤ r ≤ n, then there are

P (n, r) = n(n − 1)(n − 2)···(n − r + 1)

r-permutations of a set with n distinct elements.

**Note: **If n and r are integers with 0 ≤ r ≤ n, then** P (n, r) = n!/ (n − r)! .**

**EXAMPLE** How many ways are there to select a first-prize winner, a second-prize winner, and a third-prize winner from 100 different people who have entered a contest?

**Solution: **Because it matters which person wins which prize, So for first prize we have 100 people now for second prize we havee 99 person left and third placed can be occupied by 98 people

Consequently, the answer is P (100, 3) = 100 · 99 · 98 = 970,200.

**EXAMPLE** Suppose that there are eight runners in a race. The winner receives a gold medal, the secondplace finisher receives a silver medal, and the third-place finisher receives a bronze medal. How many different ways are there to award these medals, if all possible outcomes of the race can occur and there are no ties?

**Solution: **The number of different ways to award the medals is the number of 3-permutations of a set with eight elements. Hence, there are P (8, 3) = 8 · 7 · 6 = 336 possible ways to award the medals.

OR

We can say that we need to choose 3 person out of 8 ,after choosing we can arrange themi.e

C(8,3)*3! = (8!/5!*3!)*3! = P(8,3)

**EXAMPLE** Suppose that a saleswoman has to visit eight different cities. She must begin her trip in a specified city, but she can visit the other seven cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities?

**Solution:**the first city is determined, So we have to find permutation for rest 7 cities but the remaining seven can be ordered arbitrarily. Consequently, there are 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 ways for the saleswoman to choose her tour.

**EXAMPLE** How many permutations of the letters ABCDEFGH contain the string ABC ?

**Solution:** Because the letters **ABC** must occur as a block, we can find the answer by finding the number of permutations of six objects, namely, the block ABC and the individual letters D, E, F, G, and H. Because these six objects can occur in any order, there are 6! = 720 permutations of the letters **ABC**DEFGH in which **ABC** occurs as a block.