SOLUTION: There are 6 people who will sit in a row but out of them, Ronnie, will always be left of Annie and Rachel will always be right of Annie. How many row arrangements are possible?

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Question 1206384: There are 6 people who will sit in a row but out of them, Ronnie, will always be left of Annie and Rachel will always be right of Annie. How many row arrangements are possible?

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
There are 6 people who will sit in a row but out of them,
Ronnie, will always be left of Annie and Rachel will always be right of Annie.
How many row arrangements are possible?
~~~~~~~~~~~~~~~~~~~~ 

Without any restrictions, there are 6! = 6*5*4*3*2*1 = 720 of all possible different permutations.


For any of these permutations, consider its subset of {Ronnie, Annie, Rachel}.


For these given permutation, there are 6 different permutations of these three girls in their three positions,
where all other 3 participating persons occupy their positions unchangeable.


So, from these 6 permutations, only one, namely (Ronnie, Annie, Rachel) does satisfy both restrictions.


It implies that among all possible 720 permutations, only one sixth of them are valid.


So, the answer to the problem's question is  720/6 = 120.

Solved.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: 120

Explanation

A,B,C = Ronnie, Annie, Rachel
D,E,F = the other 3 people

A must be to the left of B.
C must be to the right of B.
In other words: we must have A__B__C where 0 or more letters will go in the blanks.

Imagine that persons D,E, and F will select numbers 1 through 6 from a hat. The selections are done without replacement. The numbers refer to seating arrangement.
1 = left most seat
6 = right most seat
Use the nCr combination formula (or Pascals Triangle) to determine there are p = 6C3 = 20 different ways to do this where order doesn't matter.
Eg: Group {1,3,4} is the same as {3,1,4}

Once persons D,E, and F find their seat, there are q = 3! = 3*2*1 = 6 ways to permute the 3 members of that group.
Think of it like a game of musical chairs except none of the chairs get taken away and everyone is able to get a seat.

Therefore, we have p*q = 20*6 = 120 ways to ensure that the ordering is A,B,C where there may or may not be a gap between Ronnie, Annie, Rachel.

A few examples of valid arrangements:
ABC,DEF
ABD,CEF
ADB,ECF

Examples that aren't valid arrangements
CAB,DEF
CAD,FEB
Both aren't valid because C is not to the right of B.
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