Probability--suzan

Assuming there is no restriction : :!7 ways

SInce 3 positions are not allowed, allowed position for him is 4, for each of these combo other can be seated in !6 ways

4*!6 = 720*4 = 2880

prachich1987 wrote:In how many ways can you sit 7 people on a bench if Suzan won’t sit on the middle seat or on either end?

a) 720
b) 1,720
c) 2,880
d) 5,040
e) 10,080

source : some online material

fairly easy one

let's get to our constraints

--- --- --- --- --- --- ---, in total 7 places and the places # 1,4 and 7 are forbidden for Suzan --> so let the people be seated in these places first, then do whatever they like

No 1 only 6 people (one is out, she's Suzan)
No 4 only 5 people (the same here)
No 7 only 4 people (the same ...)

So we've got --> 6*4*3*5*2*1*4 OR -6-__ -4-__ -3-__ -5-__ -2-__ -1-__ -4-

2,880

Thanks guys
The OA is indeed C

Usually I'm ok with these but this one threw me for a loop. Could anyone explain the mistake in my approach?

Take the unrestricted combinations and subtract out the restricted combinations

Unrestricted = 7! = 5040 possible combinations

Restriction 1: She won't sit in the first seat.
How many ways can she sit in the first seat? 1*6*5*4*3*2*1 = 6! = 620
Same for middle seat, end seat. So out of 5040 possible combinations, 3*620= 1860 are restricted.

5040 - 1860 = 3180

Thanks a lot

721tjm wrote:Usually I'm ok with these but this one threw me for a loop. Could anyone explain the mistake in my approach?

Take the unrestricted combinations and subtract out the restricted combinations

Unrestricted = 7! = 5040 possible combinations

Restriction 1: She won't sit in the first seat.
How many ways can she sit in the first seat? 1*6*5*4*3*2*1 = 6! = 620
Same for middle seat, end seat. So out of 5040 possible combinations, 3*620= 1860 are restricted.

5040 - 1860 = 3180

Thanks a lot

you are on right track

6!=720, 3*6!=2,160
5,040-2,160=2,880

Haha great... that came after I checked my math several times. Thanks.