M08-30

We can solve this question in the following way :

We have total 5 children out of which 2 are siblings and 3 are not, so if we first arrange 3 children, then we get = 3!

If these 3 children are seated, then we have 4 places remaining for those 2 siblings in which they cannot seat together

_ child 1 _ child 2 _ child 3 _ = so we have 4 places remaining, so 4P2

No. of ways in which 2 siblings do not sit together = 3! AND 4P2 ----> (AND = Multiplication, OR = Addition)
= (3*2*1) * (4*3)
= 6 * 12
= 72 ways

Hence, option C is correct

*If these 3 children are seated, then we have 4 places remaining for those 2 siblings in which they cannot seat together*

Please, could you explain why we have 4 places remaining while 3 out of 5 are occupied ?

The siblings can occupy positions either between the children labeled 1, 2, and 3, or at either end of this group of children:

__ (child 1) __ (child 2) __ (child 3) __

Hi Bunuel!
Is it possible to share the solution as per Fundamental Counting principle method?Bunuel

 
­The Fundamental Counting Principle states that if an event has x possible outcomes and a different independent event has y possible outcomes, then there are xy possible ways the two events could occur together. We do not have two different independent events in this question, so not sure how you would apply this here.

Bunuel

It may sound stupid, but iam going to ask it anyway. When you say 2 sibilings, out how many people do 2 sibilings exist?

I thought 2 sibilings consist of 4 people: Sibiling-1=(brother,sister), and Sibiling 2 (brother,sister). That is why i did the following: 5!= 120 and then, 2 sibilings=4 people compress it in two plus the last one in the group, i got 3!= 6 and 6* 2!= 12.

But then, when i subtracted the 12 from the 120 (120-12)= 108 was not among the answer choices.

My approach is good, but it seems the way i counted/interpreted 2 sibilings was wrong.


Thanks in advance!­

­The term "two siblings" refers to two people. Each sibling is one individual, so when you say "two siblings," you are talking about two individuals who are brothers, sisters, or one of each.

­Total number of ways to arrange the 5 children = Total number of ways to have the siblings sit together + Total number of ways to have the NOT sit together

Based on the above;
Total number of ways to have the siblings NOT sit together = ­Total number of ways to arrange the 5 children - Total number of ways to have the siblings sit together

Hence
Total number of ways to arrange the 5 children
5! = 120

Total number of ways to have the siblings sit together
If we have the siblings sit together, then we should take that as 1 position, which means that we have four population to arrange in 4! ways
Therefore 4! = 24
We can also arrange the siblings in 2! ways in their respective positions in the general population. That is, sibling A might be placed before sibling be or B before A
Therefore, total number of ways to have the siblings sit together = 4!*2! = 24*2 = 48

Recall;
Total number of ways to have the siblings NOT sit together = ­Total number of ways to arrange the 5 children - Total number of ways to have the siblings sit together

Therefore
Total number of ways to have the siblings NOT sit together = 120 - 48
= 72

Answer choice; C

I like the solution - it’s helpful.