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Arrangements and Selections With Repetitions

Arrangements and Selections With Repetitions. How many ways to arrange the six letters b, a, n, a, n, a ?. Six places: __ __ __ __ __ __. places to put the a’s: __ a, a, __ __ a. places to put the n’s: n, a, a , __ n, a.

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Arrangements and Selections With Repetitions

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  1. Arrangements and Selections With Repetitions How many ways to arrange the six letters b, a, n, a, n, a ? Six places: __ __ __ __ __ __ places to put the a’s: __ a, a, __ __ a places to put the n’s: n, a, a, __ n, a places left to put the b: n, a, a, b, n, a Total: Tucker, Section 5.3

  2. = 0! =1 General Formula The number of arrangements of n objects chosen from m types, with rj of type j, where , is P(n; r1, …, rm) = Tucker, Section 5.3

  3. Alternate Proof Think of b, a, n, a, n, a as 6 different letters, so 6! ways to arrange them. But then, it doesn’t matter which of the three a’s comes in which order, so have to divide by 3!. Similarly, have to divide by 2! for the two n’s. And lastly, have to divide by 1! for the b, giving: This generalizes to the same formula of P(n; r1, …, rm) = Tucker, Section 5.3

  4. Selections with Repetition How many ways are there to select 6 pieces of fruit if the choices are apples, oranges, and bananas? X X X X X X 2 apples, 3 oranges, 1 banana X X X X XX 5 apples, no oranges, 1 banana X X X X X X 4 apples, 1 orange, 1 banana Think of 6 X’s split in 2 (one less than the number of types) places. So we need to choose 6 out of 8 possible places to put the X’s. Answer: Tucker, Section 5.3

  5. Selections with RepetitionGeneral Formula The number of selections with repetition of r objects chosen from n types of objects is C(r + n –1, r). Proof: X X X … X … X … X … X There are rX’s, and we have to choose where to place n – 1 slashes to group them into the n different types, and there are C(r + n –1, r) ways to do this. Tucker, Section 5.3

  6. Example How many ways are there to pick exactly 10 balls from a (large*) pile of red, blue and purple balls, if there must be at least 5 red balls? First set aside the 5 red balls you need (only 1 way to do this). Then pick 5 more balls from the three colors: C(5 + 3 –1,5) = 21 ways to do this. *I.e. at least 10 of each color. Tucker, Section 5.3

  7. Example How many ways are there to pick exactly 10 balls from a large pile of red, blue and purple balls, if there must be at most 5 red balls? Notice that ‘at most 5 red balls’ is the complement of ‘at least 6 red balls’, so if we count the total number of ways to pick 10 balls, and then subtract the number of way to get at least 6 red balls, this will give us the number of ways to get at most 5 balls. Pick 10 balls: C(10 + 3 – 1, 10) = 66 ways total. Pick at least 6 red balls (ie set aside 6 red balls, then pick 4): C(4 + 3 –1, 4) = 15. Answer: 66 – 15 = 51 ways to choose at most 5 red balls. Tucker, Section 5.3

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