Since 3 members want to sit together, we'll calculate total ways for it and then for remaining seats in each of its case, we'll see in how many ways can those rest of 5 people wants to sit.
The total number of ways those 8 people can sit is 3,991,680 ways.
Given that:
- Total 14 seats are there in a row in a doctor's waiting room.
- 8 people are waiting to be seated.
- 3 people are of same family and want to sit together.
To find:
In how many ways can those people sit?
Number of ways in which 3 family members can sit:
Since 3 people wants to sit together, then suppose seats are numbered 1,2,3,...,14.
Then those 3 people can choose seats as:
1,2,3
2,3,4
...
...
12,13,14
Total 12 ways.
In each way, those three can sit on those 3 chairs in any order. The number of ways 3 people will arrange themselves is 3! = 6 ways.
Thus:
Total ways those 3 people can sit in = 12 times 6 = 72 ways.
( i did multiplication since for each of 3 seat chosen they were sitting in 6 ways, and this repeated 12 times would end up on 12 times 6 = 72 ways).
Number of ways those remaining 5 people can sit on remaining 11 seats:
On remaining seats (14-3 = 11), we have 5 people to sit.
Those people can choose any seat they want to sit on.
Total ways(combinations) they together choose 5 seats is given by: [tex]^11C_5 = 462 [/tex]
For each of those choice of seats, they can arrange themselves in 5! = 120 ways.
Thus, those 5 people can sit in 120 times 464 = 55440 ways.
Since on each choice of those 3 family members , we have 55440 ways of 5 people sitting on remaining 11 seats, thus in total we have:
55440 times 72 = 3,991,680 ways.
Thus, in total, those 8 people can sit in 3,991,680 number of ways in that row of the doctor's waiting room.
Learn more about combinations here:
https://brainly.com/question/10618762
