Permutation questions.. (explain only)?

Question

Q1) In how many different ways can one make a first, second, third or fourth choice among 12 firms leasing computer equipment?

The solution:

12P4 = 11880.

But I don't understand why it is 12P4...

Q2) In how many ordered ways can a television director schedule six different commercials during the six time slots allocated to commercials during the telecast of the programme "Under One Roof"?

The solution:

6! = 720...

I don't understand...

Answer 1

12P4 = 12 ! / (12-4)! = 12 ! / 8!

Choice ACBD is different from ABCD , so it is permutation and not combination. The ordering has importance.

In combination (selection) ABCD and ACBD are the same.

This is the same as ,"In how many ways can he permute 6 letters? "

6 x 5 x 4 x 3 x 2 x 1 =720

Answer 2

Q1) In how many different ways can one make a first, second, third or fourth choice among 12 firms leasing computer equipment?

The formula of 12P4 is a bit complicated and counter intuitive because it is based on factorials, if it confuses you just go back to simple counting arguments: put yourself in the place of the manager choosing. You consider the list of 12 items longly, and choose one among 12, leaving 11 for your second choice. You consider the list of 11 items and choose one, leaving 10 for the third choice, after which you will be left with 9 items for the fourth and last choice: all in all there are 12 * 11 * 10 * 9 = 11 880 choices that you can do.

Q2) In how many ordered ways can a television director schedule six different commercials during the six time slots allocated to commercials during the telecast of the programme "Under One Roof"?

Use the same argument: you ponder your list of 6 commercials till you have chosen the perfect one for the first slot in your programme: you are down to 5 for the next slot, 4 for the thirs slot, 3 for the fourth, and so on. In total there are  6 * 5 * 4 * 3 * 2 * 1 = 6! = 720 possible ways to arrange the 6 commercials in the 6 slots.

If it confuses you, try to enumerate the combinations with only 2 commercials A,B in 2 slots, and check that there are exactly 2 of them: AB and BA. Should you be given a third commercial C and a third spot,

from each combination you can get three new ones depending of where you insert C: AB yields CAB,ACB and ABC while BA yields CBA, BCA and BAC, summing to six (3*2=6) of them: ABC,ACB,BAC,BCA,CAB,CBA.

Should you be given a fourth commercian D and a fourth slot, from each combination of three commercial you can obtain four distinct combination of four commercial depending of where you insert D: e.g. ABC yields DABC, ADBC, ABDC and ABCD; ACB yields DACB, ADCB, ACDB, ACBD and so on: in total you get 4 * 6 combinations, which is really 4*3*2*1 = 4! combinations.

Hope it helps!

Answer 3

Q1. Because you're choosing 4 from 12 and the order is important. That is, 1st is IBM, 2nd is HP..., is different from 1st is HP, 2nd is IBM...

Q2. It's 6P6 which is 6!, again because order matters. For example, 1st is an IBM commercial...