Counting principle, permutations, or combinations

[mathwhizwannabe]

At a bus stop, 5 people enter a bus that has only 3 empty seats.

a. In how many different ways can 3 of the 5 people occupy these empty seats?

b. If Mrs. Costa is one 1 of the 5 people, what is the probability that she will not get a seat?

c. If Ann and Bill are 2 of the 5 people, what is the probability that they both will get seats?

My guesses:

a. 60 (5P3)
b. 2/5 (3/5 is the probability that she will get a seat-- I think)
c. 3/10 (3/5, the probability that Ann will get a seat, times 2/4, the probability that Bill will get a seat after Ann is seated)

I am not sure whether I did these right. Can anyone help?

 

[Gene]

a) nope, if joe gets seat one and jim gets seat two its the same as jim gets seat one and joe gets seat two so it is combinations, not permutations. C(5,3)

b) Good.

c) Also.

 

[soroban]

Hello, mathwhizwannabe!

At a bus stop, 5 people enter a bus that has only 3 empty seats.

a. In how many different ways can 3 of the 5 people occupy these empty seats?
My answer: \(\displaystyle P(5,3)\,=\,60\)

I agree . . . **
The five people are distinguishable (they have different names)
\(\displaystyle \;\;\)and the three seats are distinguishable (they have different locations).

b. If Mrs. Costa is one 1 of the 5 people, what is the probability that she will not get a seat?
My answer: \(\displaystyle \frac{2}{5}\;(\frac{3}{5}\) is the probability that she will get a seat-- I think)

I agree . . . I did it The Long Way . . . just to check.

There are 60 possible seatings.

If Mrs. Costa is not seated, the three seats go to the other four people.
\(\displaystyle \;\;\)There are: \(\displaystyle P(4,3)\,=\,24\) ways.

\(\displaystyle P(\text{Mrs. Costa not seated})\:=\:\frac{24}{60}\:=\:\frac{2}{5}\)

c. If Ann and Bill are 2 of the 5 people, what is the probability that they both will get seats?
My answer: \(\displaystyle \frac{3}{10}\)
\(\displaystyle (\frac{3}{5}\), the probability that Ann will get a seat,
times \(\displaystyle \frac{2}{4}\), the probability that Bill will get a seat after Ann is seated)

Again, I checked this The Long Way . . .

There are 60 possible seatings.

If Ann and Bill are seated, they can be seated in \(\displaystyle P(3,2)\,=\,6\) ways.
\(\displaystyle \;\;\)The third seat can be given any of the remaining 3 people: \(\displaystyle 3\) ways.
\(\displaystyle \;\;\)Hence, there are: \(\displaystyle \,6\,\times\,3\:=\:18\) ways for Ann and Bill to be seated.

\(\displaystyle P(\text{Ann & Bill get seats})\:=\:\frac{18}{60}\:=\:\frac{3}{10}\)

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** .Gene, you answered this question:
. . . "In how many ways can 3 people be chosen?"

 

[pka]

** .Gene, you answered this question:
. . . "In how many ways can 3 people be chosen?"

Of course it makes no difference to the answers in parts b & c whether we use permutations or combinations. If I were grading part a of this question , I would expect Gene’s answer. Reason being, taken by itself the problem belongs to a whole class of problems known as ‘occupancy problems’. From the traveler’s point of view, it hard to see how the seats are different. So we are just selecting three to be seated.