<p>In how many different ways can 5 people arrange themselves in the 5 seats of a car for a trip if only 2 of the people can drive?</p>
<p>(A) 12
(B) 15
(C) 26
(D) 48
(E) 120</p>
<p>Answer is D and that is what I got because I simply guessed since I knew it wouldn't be 120 lol. But how do you do this mathematically?</p>
<p>2<em>4</em>3<em>2</em>1 = 48.</p>
<p>Just multiply the number of people left for each seat.</p>
<p>2 possible for the driver's seat, 4 for the next seat, then 3 are left for the next seat, and so on.</p>
<p>when do you add when it's a probability question?</p>
<p>Yay! It's math!</p>
<p>Since we have two options for the driver, this leaves us 4 people for 4 seats. This is simple permutation of 4!. So, the answer is 2*4! = 48 ways.</p>
<p>For the question: "when do you add when it's a probability question?"</p>
<p>You add if you're dealing with words "and/or." Now, look at this modification of your problem.</p>
<p>There are 5 people ---- Adam, Brent, Cody, Dan, and Eric ---- in the car. Only Adam has license to drive. Now, because Brent, Cody, Dan, and Eric are crazy teens, all of them like to seat on the passenger seat. If they are allowed to put two or one person in the passenger seat and only one person on other seats, how many ways can you put people if there are 3 backseats? (by two people on one seat, you somehow squeeze two people in there and hope that cop won't catch them)</p>
<p>Solution:</p>
<p>This time we're adding probabilities. Now, we have to get rid of Adam. There is only one Adam so it's 1*something. Let's deal with one person on the passenger seat. </p>
<p>This person can be Brent, Cody, Dan, or Eric. So that's 4 ways. Now, for the rest of three seats, this is simply 3!. As a result, the answer is 1<em>4</em>3! = 24 ways.</p>
<p>Let's deal with two people on the passenger seat.</p>
<p>Two people can be either (let's use letters) BC, BD, BE, CD, CE, or DE. So there are 6 ways. Now, this leaves 2 people to place on the 3 empty backseats. If you call these two leftover people as Brent and Eric, here is what can be possible:</p>
<p>B _ C
_ B C
B C _</p>
<p>So, the answer is 1<em>6</em>3 = 18 ways.</p>
<p>The answer is 24+18 = 42 ways.</p>
<p>42 isn't even an option though..</p>
<p>LOL hahaha ^^</p>
<p>Am I missing something? or am I correct?</p>
<p>omgosh i know!! hahaha</p>
<p>okay so there is always going to be a driver (2 choices for that)</p>
<p>think of a normal car which can seat 5 people
4 3 2
1 2</p>
<p>JOE MARY SUE PAT BOB MARGE are all going for a car ride in a 5 seater.
(Marge and Bob are the only ones who can drive the car - save 2 seats for the driver)
left rear seat - 4 people can sit there (either Marge or Bob (the one not driving obviously), Joe, Mary, Sue, Pat.
since one person is already in a seat (lets say Joe), only 3 can sit in the middle chair (Mary Sue Pat are left!)..with Mary taking the seat.
leaving 2 choices for people for the rear right (Sue and Pat), Sue takes the seat
shot guns left, with only Pat left! </p>
<p>4 (rear left) x 3 (rear middle) x 2 (rear right) x 1 (shotgun) x 2 (driver =
48!</p>
<p>haha and only got a 640 on math sigh.</p>
<p>
[quote]
Now, look at this modification of your problem
[/quote]
</p>
<p>Dont you guys know how to read?</p>
<p>shouldn't it be:</p>
<p>2 x 3 x 3 x 3 = 54 but that isn't an option. so no.</p>
<p>oh wait:</p>
<p>2 x 3 x 2 x 1 = 12 and that is an option, so the answer is A.!!!!!!</p>
<p>The first seat can have only 2 people
The remaining 3 seats:
seat 1: any of the remaining 3 people
seat 2: either of the remaining 2 people
seat 3: takes the only remaining person</p>
<p>multiply the numbers together and you get 12!!!</p>
<p>but how is the answer D? wait, D is a multiple of 12. so I must have been on the right track but with a mistake in the middle. someone correct me.</p>