0
$\begingroup$

A computer monitor manufacturer produces 7 different models, all with a different selling price. Suppose that 4 different customers (call them A, B, C and D) will each purchase one of the monitors.

I just want to check my answers. If this is right or wrong.

In how many ways may this be done assuming that no two customers purchase the same model?

I did a sample space here. i assume that no two customers purchase the same model so i did like this . Customer A can purchase 7 different models. B can purchase 6 different models. C can purchase 5 different models. D can purchase 4 different models. Multiply them then i got 840 ways to do it.

In how many ways may this be done assuming that different customers may purchase the same model?

For this one, I got 5040 ways to do it because I 7!.? i honestly i am confuse in this part.

$\endgroup$
  • $\begingroup$ Same reasoning as before. First customer has $7$ choices, so does the second, and so on. Hence $7^4=2401$ $\endgroup$ – lulu Sep 15 '16 at 1:28
  • $\begingroup$ $7!$ is wrong. $A$ can purchase any one of seven models, $B$ can purchase any one of seven models,$C$ can purchase any one of seven models,and $D$ can purchase any one of seven models. So the answer is $7*7*7*7=2401$. $\endgroup$ – астон вілла олоф мэллбэрг Sep 15 '16 at 1:30
  • $\begingroup$ I really thank you guys for explaining it to me very well! $\endgroup$ – Tenji Kiriyama Sep 15 '16 at 1:45
1
$\begingroup$

Yes, I believe you are correct. For those saying it is incorrect, note that Tenji says

...assuming no two customers purchase the same model...

Also, the reason you got 5040 is that $7!$ is actually $7*6*5*4*3*2*1$, whereas you wanted $7*6*5*4$.

$\endgroup$
  • 1
    $\begingroup$ actually the part where ...assuming no two customers purchase the same model... is the 7 * 6 * 5 * 4 = 840 ways and actually got that answer before posting here... what i dont get and they answered is the question: In how many ways may this be done assuming that different customers may purchase the same model? they answered 2401 $\endgroup$ – Tenji Kiriyama Sep 15 '16 at 1:42
  • $\begingroup$ ah. never mind then. anyway, did the explanation about the factorial error help at all? lol $\endgroup$ – TrojanByAccident Sep 15 '16 at 2:09
  • $\begingroup$ They did and it make sense! I also thank you for trying to help me! Combinations and permutations word problems really confuse me. $\endgroup$ – Tenji Kiriyama Sep 15 '16 at 2:16
  • $\begingroup$ :D No problem! They can be quite confusing at first. $\endgroup$ – TrojanByAccident Sep 15 '16 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.