1. Simple Arrangements & Selections

2. Combinations & Permutations A permutation of n distinct objects is an arrangement, or ordering, of the n objects. An r-permutation of n distinct objects is an arrangement using r of the n objects. A r-combination of n objects is an unordered selection, or subset, of r of the n objects.

3. Notation P(n, r) denotes the number of r-permutations of n distinct objects. C(n, r) denotes the number of r-combinations of n distinct objects. It is spoken “n choose r”. The C(n, r) are known as “binomial coefficients” (for reasons that will become clear later).

4. Formula for P(n, r) By the product principle, P(n, 2) = n(n - 1); P(n, 3) = n(n - 1)(n - 2); … P(n, n) = n! n choices for the 1st position (n - 1) choices for the 2nd position, …, 1 choice for the n th position. P(n, r) = n(n - 1)... (n - (r - 1)) = n(n - 1)... (n - (r - 1)) (n - r)!/(n - r)! = n!/(n - r)!

5. Formula for C(n, r) All r-permutations can be counted by the product rule: 1. pick the r elements to be ordered: C(n, r) 2. Order the r elements: P(r, r) = r! That is, P(n, r) = C(n, r)P(r, r) = n!/(n - r)! Thus, C(n, r) = n!/(r!(n - r)!) C(n, r) = C(n, n - r). Give a counting argument for this.

6. Example How many ways can 7 women & 3 men be arranged in a row, if the 3 men must be adjacent? Treat the men as a block. There are C(8, 1) ways to place the block of men among the women. There are P(3, 3) ways to arrange the men. There are P(7, 7) ways to arrange the women. By the product rule, there are (8)(3!)(7!) ways.

7. Example How many ways are there to arrange the alphabet so that there are exactly 5 letters between a & b? Pick the position of the left letter of a & b. (This forces the position of the other letter.) Pick the order of a & b: 2. Arrange the other letters: P(24, 24). By the product rule: C(20, 1)(2)P(24, 24).

8. Example How many 6-digit numbers without repetition are there so that the digits are nonzero, and 1 & 2 do not appear consecutively? It is simpler to do this indirectly: There are P(9, 6) ways to arrange 6 nonzero digits without repetition. Subtract the number of ways to arrange the digits so that 1 & 2 are consecutive:

9. There are C(5, 1) ways to pick the leftmost position where the 1 or 2 go. There are 2 ways to arrange 1 & 2. There are P(7,4) ways to arrange the other letters. The number of ways to do this is: P(9, 6) - C(5, 1)(2)P(6, 4) ways to do this.

10. Example How many ways are there to arrange the letters in the word MISSISSIPPI? Since the letters are not distinct, the answer is less than P(11,11) = 11! Pick the 1 position where the M goes: C(11,1) Pick the 4 positions where the Is go: C(10,4) Pick the 4 positions where the Ss go: C(6,4) Pick the 2 positions where the Ps go: C(2,2) There are C(11,1) C(10,4) C(6,4) C(2,2) ways.

11. Example How many committees of 4 people can be chosen from a set of 7 women and 4 men such that there are at least 2 women?

12. There are at least 2 women? If we pick 2 women, then pick 2 more people without restriction, we get: C(7, 2)C(9,2). This is wrong; certain committees are counted more than once. See the tree diagram of this use of the product rule.

13. The Set Composition Rule When using the product rule, the elements of each component must be distinct. In the proposed count, we cannot always tell which women were picked in the 1st phase, & which were picked in the 2nd phase.

14. To fix this: Use the addition rule The set of committees can be partitioned into: those with 2 women: C(7, 2)C(4, 2) those with 3 women: C(7, 3)C(4, 1) those with 4 women: C(7, 4)C(4, 0) Thus, there are C(7, 2)C(4, 2) + C(7, 3)C(4, 1) + C(7, 4)C(4, 0) such committees.

15. Example How many ways are there to arrange As & Us such that the 3rd U appears as the 12th letter in a 15 letter sequence? In the 1st 11 letters, U appears exactly 2 times. Using the product rule: Pick the positions of the 1st 2 Us: C(11, 2) Pick the 3 letters that follow the 12th: 2 3.