Math 2001 Homework #10 Solutions


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1 Math 00 Homework #0 Solutions. Section.: ab. For each map below, determine the number of southerly paths from point to point. Solution: We just have to use the same process as we did in building Pascal s triangle: mark a to count the paths coming directly out of, then below that at each vertex add up all the paths coming into that vertex. For part (a) we obtain 8 paths, and for part (b) we obtain 0 paths as shown Section.:. For each map below, determine the number of southerly paths from to that pass through. Solution: The strategy here is to first figure out how many paths there are from to, then figure out how many paths there are from to. We then multiply these possibilities (since once I get to, it doesn t matter how I got there when I start heading out to ). In part (a) we get 0 paths from to using the standard Pascal s triangle (since it s a regular grid) which are colored green. Then we start over in red and get another 0 paths from to again using Pascal s triangle. So in total there are 00 paths from to that pass through. In part (b) we get paths from to colored in green. Then starting over in red we get paths from to. So in total there are southerly paths from to that pass through.
2 0 0. Section.:. social studies test consists of ten multiplechoice questions, each with five possible choices; ten True/False questions; and ten matching questions in which the ten correct answers are given and are to be matched with the appropriate question. (a) How many ways can each section of the test be randomly completed if, in the matching section, each answer is used exactly once? Solution: For the multiplechoice portion, there are ten questions and five choices for each question, so there are 0 ways to answer it. For the true/false question, there are ten questions and two choices for each question, so we get 0 ways to complete this. Finally for the matching question, since each answer must be used only once, we have 0! ways of permuting the answers and pairing with the questions. (b) How many ways may the entire test be randomly completed? Solution: Just multiply the number of ways to complete each portion: that s. Section.: ! = 0 0 0!. (a) How many different license plates are possible if each is to contain three letters followed by three digits? Solution: There are choices for each of the first three slots, then 0 choices for each of the next three slots, so a total of 0 plates. (b) How many different license plates are possible if each is to contain three letters and three digits in any order? Solution: To build such a license plate, we would have to first decide which three of the six slots will be letter slots (and all the rest will be number slots). This is the same as the number of subsets of size from the six element set {,,,,, } where the digits represent which slot we re in; so there are ( ) ways to do this.
3 Once we ve decided which slots have letters and which have numbers, the answer is the same as before: 0. So in total we have ( ) 0 possible license plates of this form.. Section.: egj. alculate: ( ) (e) = = =,88, (Here I took a 7 out of to get 7, the out of 8 to get, and the out of to get.) (g) ( ) = = 8 =,7,80. Here I factored =, 0 =, 7 =, =, ( ) (j) = = = 8, Section.: ace. =, and =. (a) Find the coefficient of x y 7 in the expansion of (x + y) 0. Solution: From the binomial theorem, it s ( ) = = 7 = 77, (b) Find the coefficient of x 7 in the expansion of (x ). Solution: The expansion will have a term like ( ) 7 x 7 ( ), so the coefficient of x 7 is ( ) ( ) 0 8 = = 8 =,. 7 (c) Find the coefficient of x in the expansion of (x + ). Solution: s before, the expansion has a term like ( ) x, so the coefficient of x is ( ) = = 7 =,78.
4 7. Suppose you remove all prime numbers from a standard deck of cards because they re unlucky, as well as removing the jacks because what s a jack? So the only faces left are {,,, 8,, 0, Q, K}, with all four suits of each. (a) How many fivecard poker hands are there with such a deck? Solution: There are cards and we need to choose, so ( ) 0 8 = = 8 8 = 0,7. (b) How many full house hands are there? (Three of one face, two of another face, such as ces of clubs, diamonds, and hearts, and s of spades and diamonds.) ( Solution: First we choose the triplet: 8 ). Then we choose the suits for the triplet: ( ( ). Next we choose the doublet: 7 ). Then we choose the suits for the doublet: ( ). The total number is 8 7 =. (c) How many straight flush hands are there? (Five cards in a row, all of the same suit, such as ce,,, 8, of hearts. (Note that {,,, 8, } and {, 0, Q, K, } both count since ce can be considered either the highest or the lowest card.) Solution: First we choose the starting card (which can be {,,, 8, }), then we choose the suit. In total there are = 0 such hands. (d) How many flush hands are there? (Five cards all of the same suit, but not in a row.) ( Solution: We need to choose the suit: ). Then we choose any five cards in that suit; there are ( ) 8 = 8 7 = ways to do this. Thus in total there are = flushes. However 0 of them are straight flushes, so only 0 of them are genuinely flushes. 8. Section.:. Refer to the maps of Exercise of Section.. For each map, what is the probability that a randomly selected southerly path from point to point passes through point? Solution: We have already done half the work here: we counted all the paths from to that pass through. To find the probability, we just need to count all the paths from to. In part (a), we fill in the usual terms of Pascal s triangle to find paths from to. n alternate way of doing it without filling in the triangle is to just use the fact that point is ten rows down and right in the middle, so it s ( ) 0 = = 7 =.
5 Either way, the probability in part (a) of a random path going through point is 00 =.7%. In part (b) we do the nonstandard Pascal s triangle to get from point to point. dding up at each vertex the numbers from all attached vertices above it, we eventually obtain paths from to. Since we already calculated that there are paths that pass through, the probability of a random path going through on the way to is = 7.%
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