math question

[ippie]

---

Members do not see this ad.

thomas has two CD'S and two tapes. in how man arrangements can he play the CD's and tapes one after the other such that each is played exactly once and one CD is not played immediately before or after the other?

=> answer : 12
any formula for this question? or do we have to count one by one?

 

[Streetwolf]

thomas has two CD'S and two tapes. in how man arrangements can he play the CD's and tapes one after the other such that each is played exactly once and one CD is not played immediately before or after the other?

=> answer : 12
any formula for this question? or do we have to count one by one?

Choose the order in which you arrange the 4 objects = 4! = 24. Consider the CD locations. There are (4 C 2) = 6 ways that the CDs could be placed into the lineup (without regard to order). Of these, 3 ways put them next to each other (CCTT, TCCT, TTCC). So 1/2 of the ways are no good. That leaves 12 good ways out of the original 24.

Another way to think about this is to consider the CDs as one object. So now you have 2 tapes and 1 CD. These can be ordered in 6 different ways (3! = 6). But recall that the CDs can be played in either order so it's really 6*2 = 12 different ways. This is the # of ways that put the CDs next to each other. You want the # of ways that do NOT. So take 24-12 = 12.

A third way is to do it position by position. There are 4 choices for the 1st selection - a tape or a CD. If it's a tape, you can only choose a CD for your next option (if you choose another tape it leaves you with 2 CDs for spots 3 and 4). So that's 2*2 (2 tapes * 2 CDs). Now you have tape-CD so you have to go with another tape, then another CD. That's 2*2*1*1. If you started with CD, you have to go with tape next. So that's 2*2. For the third choice you can select either tape or CD. So you have 2*2*2. Then the last choice is whatever you didn't choose, so you have 2*2*2*1. So that's 4 + 8 = 12.