Given 5 different green dyes, 4 different blue dyes and 3 different red dyes. How many combinations of dyes can be chosen taking atleast 1 green and 1 blue dye?. (A.4096 B.3255 C.3720)

Question

ganeshie8 · Answer

Assuming the dyes in a particular group are distinguishable,
the total number of combinations of dyes with out any restrictions would be $2^{12}$.

debpriya · Answer

Now we will subtract the combination of no green and no blue dye ? Will we also subtract 1 blue and no green and 1 green and no blue ?

ganeshie8 · Answer

Yes. Out of those $2^{12}$ total possible combinations, 
$2^{7}$ will not have any green,
$2^8$ will not have any blue,
$2^3$ will not have both green and blue.

debpriya · Answer

Thank you so much :)

ganeshie8 · Answer

Np :)

debpriya · Answer

The answer is coming out to be 3704 :(

ganeshie8 · Answer

I'm getting 3720

ganeshie8 · Answer

I know the mistake you have done

debpriya · Answer

Oh I must have done a calculation mistake then. We have to subtract  the three cases from 2^12 right ??

ganeshie8 · Answer

That is the mistake

ganeshie8 · Answer

Let me ask you a question

ganeshie8 · Answer

How many combinations have no green ?

debpriya · Answer

2^7 ?

ganeshie8 · Answer

Yes, out of those 2^7 that have no green, 
will there be few combinations that have no blue too ?

debpriya · Answer

Yes yes

ganeshie8 · Answer

Good. Another question

debpriya · Answer

Did I double count it ?

ganeshie8 · Answer

How many combinations have no blue?

debpriya · Answer

2^8 ?

ganeshie8 · Answer

Yes, out of those 2^8 that have no blue, 
will there be few combinations that have no green too ?

debpriya · Answer

Yes

ganeshie8 · Answer

How many are they ?

ganeshie8 · Answer

How many combinations are there that don't have both green and blue ?

debpriya · Answer

2^3

ganeshie8 · Answer

As you can see, we have counted them in both "no green" and "no blue"

so we must subtract one copy

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=2%5E12+-+(2%5E7+%2B+2%5E8+-+2%5E3)

debpriya · Answer

Oh!! So we won't include 2^ 3  ?

ganeshie8 · Answer

We must include 2^3 only once. Not twice.

ganeshie8 · Answer

Since we have included the common 2^3 combinations in both "no green" and "no blue",
we are subtracting it once

ganeshie8 · Answer

Lets do a quick example maybe

debpriya · Answer

Sure please

ganeshie8 · Answer

Look at below venn diagram 
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