The Carson family will purchase three used cars. There are

[vinny4nyc]

Hi
I understand the explanation given for the Slot method.I however approached the question in a different manner and got an absurd answer (not listed in the choices). Where did my thinking go wrong:

The possible ways of selecting car models was
A A A
A A B
A B B
B B B
since order was not important.

Within each group, say A A A ,he could have chosen the colour for first car in 4 ways,2nd car in 3 ways and 3rd car in 2 ways.
That is 24 possible combinations.

Extending that to all 4 possible groups 24 * 4 = 96.

Where did I go wrong here?

Regards
Vinayak

[tim]

you're overcounting. for instance, once you assign all three cars as A, your strategy will assign RGB as one color combination, but also RBG, GBR, and several others, all of which give you the same three cars!

[vinny4nyc]

Thanks Tim.

[tim]

my pleasure!

[kunal.choudhary]

Alternate approach:
We have two models of cars : A and B
We have to select 3 cars. SO according to the models the combinations will be as follws:
AAB
BBA
AAA
BBB
(remember order is not important)
Now for each case we will select different colors :-
AAB = 4C2 (selecting 2 colors out of 4 for 2 cars of model A) * 2C1 (Selecting one color, out of the remaining 2 colors, for model B) = 6*2 = 12

BBA = 4C2 * 2C1 = 6*2 = 12

AAA = 4C3 (select 3 out of 4 colors) = 4

BBB = 4C3 = 4

Now adding up the values for the 4 cases we get 12+12+4+4 = 32

[tim]

cool; thanks!

[shah.tanmay.b]

bansalm : thanks for the help.. great method...

[jlucero]

Glad we're helping each other out.

[deem432]

I am struggling with this question. I don't understand why you need to divide by 3!. I have re-read the postings and something just isn't clicking.. please help!

[RonPurewal]

I am struggling with this question. I don't understand why you need to divide by 3!. I have re-read the postings and something just isn't clicking.. please help!

when you just multiply numbers of possibilities together, "order matters". in other words, the first, second, and third choices are considered distinct.
so, for instance, "red A, black A, blue A" and "black A, blue A, red A" are mistakenly considered different.

there are 3! = 6 ways to scramble any set of three specific things, so any of the possible sets of three cars is going to be counted 6 times, rather than just once. to correct for this multiple-counting, we divide.

[AnuH387]

4C3 ( from the first group)+ 4c3 (second group) +4C2 *3c1( two from the first group and 1 from second from the remaining 3 colors )+4c2 *3C1

=4+4+12+12=32

Regards,
Anu

[RonPurewal]

i assume you're trying to ask ... something?
(this discussion has been inactive for almost a year, so it would be very strange if someone just showed up and randomly posted a 'solution'.)

do you have a question? if so, please clarify.

thanks.

[AnuH387]

Oh I did not realize that .. Sorry ..


Anu

[RonPurewal]

ok.

[NSH109]

"....Model A and Model B, each of which is available in four colors: blue, black, red, and green..."
Sorry but I do not understand the wording of this problem. So if a product comes in 4 colors (or available in 4 colors), there are only exactly 4 items of that product? since I did not know how to make of this, I struggled with this problem for quite a long time.