1. Copyright © Peter Cappello 2011 Simple Arrangements & Selections

2. Copyright © Peter Cappello 2011 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. An r-combination of n objects is an unordered selection, or subset, of r of the n objects.

3. Copyright © Peter Cappello 2011 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. Copyright © Peter Cappello 2011 Formula for P(n, r) By the product rule, P(n, 1) = n; 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. Copyright © Peter Cappello 2011 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)!)

6. Copyright © Peter Cappello 2011 C(n, r) = C(n, n - r). Give a counting argument for this.

7. Copyright © Peter Cappello 2011 Example How many ways can 7 women & 3 men be arranged in a row, if the 3 men must be adjacent?

8. Copyright © Peter Cappello 2011 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. 1.There are P(8, 8) ways to arrange the 7 women + 1 block. 2.There are P(3, 3) ways to arrange the men. By the product rule, there are 8!3! ways.

9. Copyright © Peter Cappello 2011 Example How many ways are there to arrange the alphabet so that there are exactly 5 letters between a & b?

10. Copyright © Peter Cappello 2011 How many ways are there to arrange the alphabet so that there are exactly 5 letters between a & b? 1.Pick the position of the left letter of a & b: C(20, 1) (This forces the position of the other letter.) 2.Pick the order of a & b: P(2, 2) 3.Arrange the other letters: P(24, 24). By the product rule: C(20, 1)P(2, 2)P(24, 24)

11. Copyright © Peter Cappello 2011 Example How many 6-digit numbers without repetition are there so that the digits are nonzero, and 1 & 2 do not appear consecutively?

12. Copyright © Peter Cappello 2011 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:

13. Copyright © Peter Cappello 2011 1.There are C(5, 1) ways to pick the leftmost position where the 1 or 2 go. 2.There are P(2, 2) ways to order 1 & 2. 3.There are P(9 - 2, 4) ways to order the other 4 digits. The number of ways to do this thus is: P(9, 6) - C(5, 1)P(2, 2)P(7, 4).

14. Copyright © Peter Cappello 2011 Example How many ways are there to arrange the letters in the word MISSISSIPPI?

15. Copyright © Peter Cappello 2011 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! 1.Pick the 1 position where the M goes: C(11,1) 2.Pick the 4 positions where the Is go: C(10,4) 3.Pick the 4 positions where the Ss go: C(6,4) 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.

16. Copyright © Peter Cappello 2011 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?

17. Copyright © Peter Cappello 2011 There are at least 2 women? If we 1.pick 2 women 2.pick 2 more people without restriction, we get: C(7, 2)C(9,2). This is wrong; some committees are counted more than once. Use a tree diagram to identify this product rule misuse.

18. Copyright © Peter Cappello 2011 The Set Composition Rule When using the product rule, each component must be selected in a distinct stage. In the incorrect counting process, we cannot tell which women were picked in stage 1, & which were picked in stage 2.

19. Copyright © Peter Cappello 2011 To fix this: Use the addition rule Partition the set of committees 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) There thus are C(7, 2)C(4, 2) + C(7, 3)C(4, 1) + C(7, 4)C(4, 0) such committees.

20. Copyright © Peter Cappello 2011 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?

21. Copyright © Peter Cappello 2011 Example Solution 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.

22. Copyright © Peter Cappello 2011 Use the product rule: 1.Pick the positions of the 1st 2 Us: C(11, 2) 2. Put As in the 9 open positions: 1 3. Put the 3 rd U in position 12: 1 4. Pick the 3 letters that follow the 12th: 2 3.