Question

How many ways can 11 Redcoat basketball players and Coach Woodburn (12 people) be arranged along the sidelines?

a) with no restrictions.

b) if Coach Woodburn must be at one end of the bench?

c) if Travis and Pat must stand together?

d) if Travis, Pat, and Woody must stand together?

e) if Travis and Pat must be kept separated?

Answer 1

a) The first in line can be any of the 12 people
the second in line can be any of the 11 remaining (1 already in line)
the third in line can be any of the 10 remaining (2 already in line) and so on.
Therefore the total number of permutations = 12 x11x10x9x8x7x6x5x4x3x2x1 = 12! = 479001600.

b) The other 11 except the coach can be arranged in 11! permutations ie the first one of 11, the second one of the 10 remaining etc.
11! = 11 x 10 x 9 x 8 x7 x6 x5 x4 x3 x2 x1 = 39916800 different ways.
Coach Woodward then has to be added to one of either ends. For each and every one of those 11! ways Coach Woodburn can sit on either the right side or the left side. Therefore the answer is 2 x 11! which equals 2 x 39916800 =79,833,600.

c. Suppose for one moment that Travis and Pat are 1 person called "TravisPat". Therefore there are 11 people to work out the permutations for. 10 regular people plus "TravisPat". As we have seen in the part (b) that the 11 people can be arranged in 11! ways. or 39916800.
HOWEVER all 11! of these permutations have Travis sitting to the right of Pat. What about the permutations when "TravisPat" is "PatTravis"? Obviously there are another 11! permutations identical to the first lot but just where Pat and Travis change seats.
Therefore the answer is 2 x 11! which equals 2 x 39916800 = 79,833,600

d. If Travis Pat and Woody stand together treat these as 1 person named TravisWoodyPat. Therefore, we have to arrange 10 people (9 regular plus TravisWoodyPat). These 10 can be arranged in 10! or 10x9x8x7x6x5x4x3x2x1 or in other words 3,628,800 permutations. However as above Travis Pat And Woody can then change their seats. There are 3! or 3x2x1 ie 6 combinations that you can order these three people's seats. These are (TPW, TWP, PWT, PTW, WPT,and WTP). Therefore the total number of permutations is 10! x 3! = 3,628,800 x 6 = 21,772,800 ways.

E. If Travis and Pat must be kept separated then the number of ways = 12! - the number of ways they stand together

= 479001600-79833600= 399,168,000.