Just to drill it further -

If you using the slotting technique ( as you pointed out in your PM) you could go ahead by this method -

1. First case (4 small and one large) -Choosing small books first - 8 ways to choose the first, 7 ways to choose the next, ... 5 ways to choose the 4th.

So (8*7*6*5) / 4! ( we divide by 4! as the order of books is not important)

and choosing one large book in 2 ways.Hence \(70*2 = 140\) ways.

2. Second case : Now slotting to choose 5 small books -(8*7*6*5*4) ways to choose the five books and divide by 5! since the order is not important.

Hence in \(\frac{(8*7*6*5*4)}{5!} = \frac{(8*7*6*5*4)}{(2*3*4*5)}= 56\) ways

And then add the two as per OR -rule

Hence a total of 196. By either method.

UNSTOPPABLE12 wrote:

Gladiator59 wrote:

Replying to a PM :

On first sight, you might fall for the trap that there is only one possibility - 4 small books and 1 large book. However, given information that Romi has space for only one large book (and assuming that one small book can replace one large book by common sense), there are actually two possibilities -

1. 4 small books out of 8 possibilities and 1 large book out of 2 possibilities -

This is in 8C4*2C1 = \(\frac{8!}{(4!*4!)}*2 = \frac{(5*6*7*8)}{(2*3*4)}*2 = 140\) ways

2. 5 small books out of 8 possibilities -This in 8C5 = \(\frac{8!}{(5!*3!)} = \frac{(6*7*8)}{(2*3)} = 56\) ways.

So a total of (1) or (2)... using the OR-RULE

=140 + 56 = 196 ways.

Hence

Option (D) is our bet.UNSTOPPABLE12, Hope it is clear now.

Let me know if not.

Perfect explanation, indeed i didnt take into account the two different scenarios, thank your for your time

Gladiator59
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Regards,

Gladi

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)