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Our last example for this section introduces the idea of a run, a notion that arises in statistics-in particular, in the detecting of trends in a statistical process. The counter at Patti and Terri's Bar has 15 bar stools. Upon entering the bar Darrell finds the stools occupied as follows: O O E O 0 0 0 EEE O O O E O, where O indicates an occupied stool and E an empty one. (Here we are not concerned with the occupants of the stools, just whether or not a stool is occupied.) In this case we say that the occupancy of the 15 stools determines seven runs, as shown: 00 E 0000 EEE O00 E O Run Run Run Run Run Run Run In general, a run is a consecutive list of identical entries that are preceded and followed by different entries or no entries at all. A second way in which five E's and 10 O's can be arranged to provide seven runs is E O 0 O E E OO E O 0 0 0 0 E. We want to find the total number of ways five E's and 10 O's can determine seven runs. If the first run starts with an E, then there must be four runs of E's and three runs of O's. Consequently, the last run must end with an E. Let x1 count the number of E's in the first run, x2 the number of O's in the second run, x3 the number of E's in the third run,..., and x, the number of E's in the seventh run. We want to find the number of integer solutions for Xi + x3 + xs +x7 - 5, X1, X3, xs, X7 >0 (3)

Our last example for this section introduces the idea of a run, a notion that arises in statistics-in particular, in the detecting of trends in a statistical process. The counter at Patti and Terri's Bar has 15 bar stools. Upon entering the bar Darrell finds the stools occupied as follows: O O E O 0 0 0 EEE O O O E O, where O indicates an occupied stool and E an empty one. (Here we are not concerned with the occupants of the stools, just whether or not a stool is occupied.) In this case we say that the occupancy of the 15 stools determines seven runs, as shown: 00 E 0000 EEE O00 E O Run Run Run Run Run Run Run In general, a run is a consecutive list of identical entries that are preceded and followed by different entries or no entries at all. A second way in which five E's and 10 O's can be arranged to provide seven runs is E O 0 O E E OO E O 0 0 0 0 E. We want to find the total number of ways five E's and 10 O's can determine seven runs. If the first run starts with an E, then there must be four runs of E's and three runs of O's. Consequently, the last run must end with an E. Let x1 count the number of E's in the first run, x2 the number of O's in the second run, x3 the number of E's in the third run,..., and x, the number of E's in the seventh run. We want to find the number of integer solutions for Xi + x3 + xs +x7 - 5, X1, X3, xs, X7 >0 (3)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Our last example for this section introduces the idea of a run, a notion that arises in statistics – in particular, in the detecting of trends in a statistical process. The counter at Patti and Terri's Bar has 15 bar stools. Upon entering the bar Darrell finds the stools occupied as follows: EXAMPLE 1.41 O O E 0 0 0 0 EEE O 0 O E o, where O indicates an occupied stool and E an empty one. (Here we are not concerned with the occupants of the stools, just whether or not a stool is occupied.) In this case we say that the occupancy of the 15 stools determines seven runs, as shown: 00 E 0000 EEE 000 E o Run Run Run Run Run Run Run In general, a run is a consecutive list of identical entries that are preceded and folowed by different entries or no entries at all. A second way in which five E's and 10 O's can be arranged to provide seven runs is E O 0 O E E 0 O E O 0 0 0 O E. We want to find the total number of ways five E's and 10 O's can determine seven runs. If the first run starts with an E, then there must be four runs of E's and three runs of O's. Consequently, the last run must end with an E. Let x, count the number of E's in the first run, x2 the number of O's in the second run, x3 the number of E's in the third run, ..., and x, the number of E's in the seventh run. We want to find the number of integer solutions for Xi + x3 + xs +x7 = 5, X1, X3, x5, x7 > 0 (3)
Transcribed Image Text:Our last example for this section introduces the idea of a run, a notion that arises in statistics – in particular, in the detecting of trends in a statistical process. The counter at Patti and Terri's Bar has 15 bar stools. Upon entering the bar Darrell finds the stools occupied as follows: EXAMPLE 1.41 O O E 0 0 0 0 EEE O 0 O E o, where O indicates an occupied stool and E an empty one. (Here we are not concerned with the occupants of the stools, just whether or not a stool is occupied.) In this case we say that the occupancy of the 15 stools determines seven runs, as shown: 00 E 0000 EEE 000 E o Run Run Run Run Run Run Run In general, a run is a consecutive list of identical entries that are preceded and folowed by different entries or no entries at all. A second way in which five E's and 10 O's can be arranged to provide seven runs is E O 0 O E E 0 O E O 0 0 0 O E. We want to find the total number of ways five E's and 10 O's can determine seven runs. If the first run starts with an E, then there must be four runs of E's and three runs of O's. Consequently, the last run must end with an E. Let x, count the number of E's in the first run, x2 the number of O's in the second run, x3 the number of E's in the third run, ..., and x, the number of E's in the seventh run. We want to find the number of integer solutions for Xi + x3 + xs +x7 = 5, X1, X3, x5, x7 > 0 (3)
and X2 + x4 + %6 = 10, X2, X4, X6 > 0. (4) The number of integer solutions for Eq. (3) equals the number of integer solutions for yı + ys + ys + y7 = 1, Yı. y3, ys. y 2 0. This number is (*+!-) = (1) = 4. Similarly, for Eq. (4), the number of solutions is (**}-) = () = 36. Consequently, by the rule of product there are 4 - 36 = 144 arrange- ments of five E's and 10 O's that determine seven runs, the first run starting with E. The seven runs may also have the first run starting with an O and the last run ending with an O. So now let w, count the rumber of O's in the first run, wz the number of E's in the second run, w3 the number of O's in the third run,..., and w7 the number of O's in the seventh run. Here we want the number of integer solutions for wi + w3 + ws + w, = 10, Wi, W3, Ws, w, > 0 and w2 + ws + w6 = 5, w2, w4, W6 > 0. Arguing as above, we find that the number of ways to arrange five E's and 10 0's, resulting in seven runs where the first run starts with an O, is (**:-)(*-') = 504. Consequently, by the rule of sum, the five E's and 10 O's can be arranged in 144 + 504 = 648 ways to produce seven runs. +1
Transcribed Image Text:and X2 + x4 + %6 = 10, X2, X4, X6 > 0. (4) The number of integer solutions for Eq. (3) equals the number of integer solutions for yı + ys + ys + y7 = 1, Yı. y3, ys. y 2 0. This number is (*+!-) = (1) = 4. Similarly, for Eq. (4), the number of solutions is (**}-) = () = 36. Consequently, by the rule of product there are 4 - 36 = 144 arrange- ments of five E's and 10 O's that determine seven runs, the first run starting with E. The seven runs may also have the first run starting with an O and the last run ending with an O. So now let w, count the rumber of O's in the first run, wz the number of E's in the second run, w3 the number of O's in the third run,..., and w7 the number of O's in the seventh run. Here we want the number of integer solutions for wi + w3 + ws + w, = 10, Wi, W3, Ws, w, > 0 and w2 + ws + w6 = 5, w2, w4, W6 > 0. Arguing as above, we find that the number of ways to arrange five E's and 10 0's, resulting in seven runs where the first run starts with an O, is (**:-)(*-') = 504. Consequently, by the rule of sum, the five E's and 10 O's can be arranged in 144 + 504 = 648 ways to produce seven runs. +1
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