There are 5 chairs. Bob and Rachel want to sit such that Bob is always

If you choose two chairs from the five, there will be only way to seat Bob and Rachel so Bob is on the left. We can choose two things from five in 5*4/2! = 10 ways.

Explanation:

We are fixing BOB sitting
B _ _ _ _ : Rachel can sit in 4 ways
_ B _ _ _ : Rachel can sit in 3 ways
_ _ B _ _ : Rachel can sit in 2 ways
_ _ _ B _ : Rachel can sit in 1 way

Total : 4 + 3 + 2 + 1
= 10 ways

IMO-C

(BR) (1 2 3 )
Let us take BR to be one unit.
Thus number of arrangements = 4!
Also, since B has to be on the left of R it cannot be arranged in 2! ways internally.

Thus IMO - Ans = 24

What am I doing wrong over here?

You're answering a different question from the one in this thread - you're solving the problem "in how many ways can the five people B, R, X, Y and Z be arranged in a row if B must be immediately to the left of R?" The answer to that question is 24, as you solved above.

The question in this thread is different for two reasons: here we only have two people, B and R, and the other three chairs are empty. So we're really arranging these letters: B, R, E, E, E, where "E" represents an empty seat. And in this question, B needs to be somewhere to the left of R, but there might be empty seats between them, so we don't want to think of them as a single unit "BR".

The answer to the problem is small enough that it's not too bad to just list all the possibilities, and if my explanation above is unclear, this should illustrate what's going on - if you do that systematically, you'll get the answer 10:

BREEE
BEREE
BEERE
BEEER
EBREE
EBERE
EBEER
EEBRE
EEBER
EEEBR

though there are much faster ways to answer the question, of course.

suppose, There are 5 fixed chairs namely A, B, C, D & E

the no. of ways to select 2 chairs among 5 chairs= \(5C2 = 10\) (answer)

all 10 cases are
AB,AC,AD,AE,BC,BD,BE,CD,CE,DE=10 (ALWAYS PLACE BOB TO THE LEFT OF RACEL NO NEED TO SUFFLE THOSE IN 2! WAYS)

@Bunuel; can you please provide a solution to this.

For this problem, I literally wrote out the scenarios (I know, not the most efficient). I also know the answer choices do not account for this solution, but can someone humor me?

There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?

_ _ _ B R
_ _ B _ R
_ B _ _ R
B _ _ _ R
_ _ B R _
_ B _ R _
B _ _ R _
_ B R _ _
B _ R _ _
B R _ _ _

So clearly, 10 possibilities. My question: why aren't the remaining blank spots factorial?

For _ _ _ B R, for example, there are 3! possible ways to arrange the people in the remaining seats. The same for the other 9 remaining seating arrangements for Bob & Rachel. So why wouldn't the answer be 60?

TIA

We would need to multiply by 3! if we were seating three people in the remaining seats, but in this particular question, the remaining seats are all empty.

Hi Ian,

Thank you so much. I guess it was my fault for assuming the three other seats were occupied. So unless the question specifically says the other seats are occupied, it is implied they are empty.

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