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Unformatted text preview: HW4 Solutions Dilip Raghavan September 28, 2008 1 Section 5.4 Problem 48. Let us first make an inventory of everything we have. The vowels: one I, one U and one O. The consonants: one N, one S, two Ts, two Rs and one C. Now, we know that the vowels must occur in alphabetical order, so let’s first put them down: I O U We also know that we must begin or end with a block of 2 Ts. We must decide whether we begin or end with them: 2 choices. TT I O U We are left with 5 letters to arrange. Let us first think about how many ways there are to distribute 5 empty slots among the letters we have arranged so far. We will then count the number of ways to arrange the letters that are left over in those empty slots. So we want to distribute 5 identical empty slots into 4 boxes: TT box 1 I box 2 O box 3 U box 4 Since we must have atleast 2 consonants between any two vowels, we must make sure that there are atleast 2 empty slots in each of boxes 2 and 3. So put two slots into each of those boxes. We have just one slot left, which we much distribute into 4 boxes: 4 ways. Now we must decide how the letters are placed in the slots: TT — I — — O — — U Since there are two identical Rs, there are 5! 2! ways to do this. So the answer is 2 × 4 × 5! 2! Problem 50. Again, let us begin by making an inventory of everything we have. The vowels: two Es, two Is. The consonants: one R, one V, one S, one T and one D. (a) Let us count the complement, which is the number of arrangements with vowels in increasing order. Firstly, the total number of arrangements is 9! 2!2! . Now if the vowels are in increasing order, let us first write them down E E I I Again, let us first distribute 5 empty slots for the 5 remaining letters into 5 boxes: parenleftBig 5 + 5 − 1 5 parenrightBig ways. Now we need to place the letter in those slots: 5! ways. So the answer is 9! 2!2! − bracketleftBiggparenleftBigg 5 + 5 − 1 5 parenrightBigg × 5! bracketrightBigg (You could also do this problem by first arranging the consonants and then distributing slots for vowels). 1 (b) Let us first arrange the vowels and then we will distribute slots for the consonants. But we must break up into four cases depending on whether the arrangement of the vowels has consecutive Es and Is or not. • Case I: no consecutive Es or Is. This means there is an I between the two Es and an E between the 2 Is. There are 2 such arrangements of the Es and Is. For example, such an arrangement might look like: E I E I Now, we are free to distribute 5 slots for the 5 consonants in any way we please into 5 boxes: parenleftBig 5 + 5 − 1 5 parenrightBig ways. Next we place the consonants in these 5 slots: 5! ways. So for this case we have 2 × parenleftBigg 5 + 5 − 1 5 parenrightBigg × 5! possibilities....
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- Spring '11
- break up, Left-handedness, ways, Disjoint sets