Permutations 2

This is a continuation of Permutations.

Consider the following problem: Lisa wants to put 5 ornaments on her mantle, but she has 12 ornaments in total she can use. How many ways can she do this? Using the Rule of Product, Lisa has 12 choices for what to put in the first position, 11 for the second, 10 for the third, 9 for the fourth and 8 for the fifth. So the total number of choices she has is 12×11×10×9×8 12 \times 11 \times 10 \times 9 \times 8 . Using the factorial notation, we write this more compactly as 12!7! \frac{12!}{7!} .

Using the same argument, we can proceed with the general case. If we have n n objects and we want to arrange k k of them in a row, there are n!(nk)! \frac{n!}{(n-k)!} ways to do this. This is also known as a kk-permutation of nn, and is denoted by Pkn P_k ^n .

Let's consider a different extension of the permutation problem. What happens if Lisa has some ornaments that are the same? If she has 2 identical cat ornaments, 3 identical dog ornaments and 3 other completely different ornaments, how many ways can they all be arranged on her mantle? In total there are 7 objects, and if we pretend they are all distinct, there are 7! 7! ways to arrange them on the mantle. For any arrangement, we can swap the pair of cats and get the same arrangement back. Also, we can move the dogs around and again get the same arrangement. How many ways can the dogs be moved around? Since the positions of the dogs are fixed, it is just the number of permutations of the dogs, which is 3!. 3!. Thus, to account for these repeated arrangements, we divide out by the number of repetitions to obtain that the total number of permutations is 7!3!2! \frac{7!}{3!2!} .

Worked Examples

1. Out of a class of 30 students, how many ways are there to choose a class president, a secretary and a treasurer? A student may hold at most 1 post.

Solution 1: There are 30 students to pick for the class president, which leaves 29 students for the secretary and 28 students for the treasurer. Hence, by the rule of product, there are 30×29×28=24360 30 \times 29 \times 28 = 24360 ways.

Solution 2: By the above discussion, there are P2730=30!(303)! P_{27}^{30} = \frac {30!}{(30-3)!} ways. While it is extremely hard to evaluate 30! 30! and 27! 27!, we notice that dividing out gives 30×29×28=24360 30 \times 29 \times 28 = 24360 .


2. How many ways can the letters in the name RAMONA be arranged?

Solution: As before, if we treat the A's as distinct from each other (say A1 A_1 and A2 A_2 ), then there are 6!=720 6!= 720 ways to rearrange the letters. However, since the letters are the same, we have to divide by 2! 2! to obtain 7202!=360 \frac {720}{2!} = 360 ways.


3. 6 friends go out for dinner. How many ways are there to sit them around a round table? Rotations of a sitting arrangement are considered the same, but a reflection will be considered different.

Solution 1: Since rotations are considered the same, we may fix the position of one of the friends, and then proceed to arrange the 5 remaining friends clockwise around him. Thus, there are 5!=120 5! = 120 ways to arrange the friends.

Solution 2: There are 6! 6! ways to seat the 6 friends around the table. However, since rotations are considered the same, there are 6 arrangements which would be the same. Hence, to account for these repeated arrangements, we divide out by the number of repetitions to obtain that the total number of arrangements is 6!6=120 \frac {6!}{6} = 120 .

Both solutions are equally valid and illustrate how thinking of the problem in a different manner can yield another way of calculating the answer.

Note by Arron Kau
7 years, 3 months ago

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