Problem Solving

I am following the Manhattan 4th edition word translations book.

I faced the following questions while doing the chapter on Combinatorics.

If seven people board an airport shuttle with only three available seats, how
many different seating arrangements are possible? (Assume that three of
the seven will actually take the seats.)

The following solution was given for the above problem.

Three of the people will take the seats (designated I, 2, and 3), and the other four will be
left standing (designated "5"). The problem is therefore equivalent to finding anagrams of
the "word" 123SSSS, where the four S's are equivalent and indistinguishable.

7!/ 4! = 210

Now consider this problem.

If three of seven standby passengers are selected for a flight, how many
different combinations of standby passengers can be selected?

At first, this problem may seem identical to the previous one, because it also involves selecting
3 elements out of a set of 7. However, there is a crucial difference. This time, the three
"chosen ones" are also indistinguishable, whereas in the earlier problem, the three seats on
the shuttle were considered different. As a result, you designate all three flying passengers as
F's. The four non-flying passengers are still designated as N's. The problem is then equivalent
to finding anagrams of the "word" FFFNNNN.

7! / (3! * 4!) = 35

My question is

How are both problems different?

Thanks for the clear solution.

But could you explain using anagram method as I had difficulty in understanding that method.

vishugogo wrote:I am following the Manhattan 4th edition word translations book.

I faced the following questions while doing the chapter on Combinatorics.

If seven people board an airport shuttle with only three available seats, how
many different seating arrangements are possible? (Assume that three of
the seven will actually take the seats.)

First select 3 people out of 7 to be seated => 7C3
Then arrange them => 3!
So the total number of arrangements = 7C3 * 3! = 7!/(3!4!) * 3! = 7!/4! = 7*6*5 = 210

vishugogo wrote: Now consider this problem.

If three of seven standby passengers are selected for a flight, how many
different combinations of standby passengers can be selected?

First select 3 people out of 7 => 7C3
That's it! You don't have to arrange them. They have asked us to select 3 people, not to arrange them.
So the total number of combinations = 7C3 = 7!/(3!4!) = 35

Cheers