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2 Answers

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i) Take A and B as a single character. Let us denote it by X. Therefore, XCDEF is our set of letters. Hence possible arrangements are 5!*2! (2! because AB can be arranged among themselves in 2! ways)

ii)There are 6 letters in total. So, 6! arrangements are possible in total. Out of those arrangements, half of them have A before B and the other half have B before A

iii)   For any of the 3! possible orders that A, B, C can come in, there are equal numbers of permutations of A, B, C, D, E, F in which A, B, C occur in the order. So for the count, we divide 6! by 3!

iv) The probability that A comes before B is 1/2. Given that A comes before B, the probability C comes before D is 1/2. There are 6! equally likely permutations. The probability of A before B and C before D is 1/2^2, so the number of "favorable" is 6!/2^2.

v)

Gluing B to the back of A and D to the back of C yields 4! = 24 different orderings in which B immediately follows A and D immediately follows C. Since the order of A and B and of C and D can be reversed, there are 4 · 24 = 96 different arrangements.

(vi) There are 5! orderings in which E is last. Hence, there are 6! − 5! = 600 orderings in which E is not last

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A) A and B are next to each other -----------------------------------------------------------> 5!*2!
B) A is before B -------------------------------------------------------------------------------------> C(6,2)*4!
C) A is before B and B is before C------------------------------------------------------------> C(6,3)*3!
D) A is before B and C is before D------------------------------------------------------------> C(6,2)*C(4,2)*2!
E) A and B are next to each other and C and D are next to each other---------> 4!*2!*2!
F) E is not last in line -------------------------------------------------------------------------------> 5*5!

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