rishabhjain13 wrote:
Hi Karishma,
Why do we need to 'unarrange' the cars, since with 8C1.6C1.4C1, we have only picked three cars and the 1st car picked can be put at any slot (1/2/3), but since we are not concerned with the arrangement, we have anyway not multiplied the expression by 3!, and hence, left it unarranged.
I'm sorry, I know I am getting confused here, but would be thankful if you could clarify this a bit further.
Thank you.
When we say 8C1*6C1*4C1, we have already arranged them in slot1, slot2 and slot 3 without even multiplying by 3!.
Here is why:
Say these are your 8 cars: BlackA, BlackB, GreenA, GreenB, RedA, RedB, BlueA, BlueB
Now, you do 8C1 and select BlackA.
Leftover cars: GreenA, GreenB, RedA, RedB, BlueA, BlueB
Of these 6, you do 6C1 and select GreenB
Leftover cars: RedA, RedB, BlueA, BlueB
Of these 4, you do 4C1 and select RedA
- Done
Now consider another case:
8 cars: BlackA, BlackB, GreenA, GreenB, RedA, RedB, BlueA, BlueB
Now, you do 8C1 and select RedA.
Leftover cars: BlackA, BlackB, GreenA, GreenB, BlueA, BlueB
Of these 6, you do 6C1 and select GreenB
Leftover cars: BlackA, BlackB, BlueA, BlueB
Of these 4, you do 4C1 and select BlackA
- Done
The two cases are different but what you got from them is the same {BlackA, RedA, GreenB}. You cannot count them twice.
The method of 8C1 * 6C1 * 4C1 is the basic counting principle in which you are automatically putting them in slot1, slot2 and slot3. For only selection, you do need to un-arrange by dividing by 3!.
Check out these posts too:
https://www.veritasprep.com/blog/2011/1 ... inatorics/https://www.veritasprep.com/blog/2011/1 ... binations/ _________________
Karishma
Veritas Prep GMAT Instructor
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