8 take any arrangement of the 21 consonants we can

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8. Take any arrangement of the 21 consonants. We can put a position in front, in between the consonants, and one at the end for a total of 22 positions. To ensure there are no pairs of vowels appearing consecutively we choose 5 positions out of the 22 in C(22,5) ways and place the vowels in these positions. Once they are in position they can be arranged in 5! ways. The consonants can be arranged in 21! ways. Our answer is C(22,5) x 5! X 21!. 9. Lay out the 14 H’s. There is only one way we can do this. Put a position in front, in between and at the end of the H’s for a total of 15 positions. To ensure that there are no consecutive tails, we choose 6 positions out of the 15 in C(15,6) ways and put a T in each position. There is only one way to arrange the T’s once they are in position. Our answer is C(15,6). 10. a. Model: selection with repetition. The number of solutions to x 1 +x 2 +x 3 +x 4 +x 5 = 28 equals the number of ways to distribute 28 identical balls into five boxes without restriction. This can be done in C(28+5-1,28) ways. b. x i >0. Each box has at least one ball in it. Distribute one ball into each box. This can only be done in one way. The remaining 23 balls can be distributed without restriction in C(23+5-1,23) ways. c. x i i . Put one ball in box 1, two balls in box 2, three balls in box 3, four balls in box 4, and five balls in box 5. This can only be done in one way. The remaining 13 balls can be distributed without restriction in C(13+5-1, 13)ways.
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12 11. Let x i = u i - 5. u i =x i + 5 -5+5 = 0. u 1 +u 2 +u 3 = 15. This is equivalent to the number of ways to distribute 15 identical balls into 3 distinct boxes with no restrictions. This can be done in C(15+3-1,15) ways. 12. For positive integer solutions, x i >0 or x i 1. We want the number of integer solutions to x 1 +x 2 +x 3 +x 4 99 1 where x i for i =1,2,3,4. The number of solutions to x 1 +x 2 +x 3 +x 4 = k with x i 1 is C(k-4+4-1, k). Summing up all the cases where k=1,2,…,99 we get C(k 1,k) k 1 99 - = . 15. We have three different teddy bears and nine identical lollipops. a. There are 4 3 ways to distribute the teddy bears, and C(9+4-1,9) ways of distributing the lollipops. There is a total of 4 3 x C(9+4-1,9) ways to distributing these without restriction. b. Each child can have at most one teddy bear. There only three teddy bears, so one child must go without. The teddy bears can be distributed to three children in 3! ways . Distribute the lollipops without restriction in C(9+4-1,9) ways. There are 4 ways to choose the child who will not receive a teddy bear. Putting these together we have 4 x 3! x C(9+4-1, 9). c. Distribute the teddies in 4 3 ways. We can ensure that each child gets 3 goodies by filling out each child with the lollipops. 16. The letters of EFFLORESCENCE , can be written as E E E E F F C C L O R S N. Put the C’s together in a box and count them like one letter. Do the same with the F’s. Consider any arrangement of the letters L O R S N CC and FF. We can put a position in front, in between and at the end of the arrangement for a total of 8 positions. To ensure that there are no consecutive E’s, we choose 4 of these positions in C(8,4) ways and put the E’s in them. The
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  • Fall '06
  • miller
  • Numerical digit, ways

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