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Simple Arrangements & Selections

Simple Arrangements & Selections 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.

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Simple Arrangements & Selections

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  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 nth 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. • By the addition rule, there are: 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 • A committee of 4 people is to be chosen from a set of 7 women and 4 men such that: • There are at least 2 women? • The committee has 2 of each sex, and Pres. & Mrs. Clinton cannot both be on the same committee?

  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 previous example, 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: 23.

  16. Characters •     •        •    •   •      •         

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