Test by Bikram | Mathematics | Test 2 | Question: 8

Question

There are three divisions of employees and each category has $5$ employees. The total number of ways a team of $8$ employees can be formed (where at least $2$ members from each division must be included in the team) is _________.

Comments

Can we solve this by

X1+X2+X3=8
 n+r-1Cr anyway?

Answer 1

(3,3,2)(3,2,3)(2,3,3)(4,2,2)(2,4,2)(2,2,4) ways and count them using combinations

Comments

Sir,

I understood what you did above but , I solved using the following approach -

Select two members from each group which is 5C2 * 5C2 *5C2.

Now we have remaining 9 people in total left . Now there is no restriction on to fill the remaining two spots - there are 9 choices * 8 choices.

Final Answer  - 5C2 * 5C2 *5C2 * 9 *8

It comes out to be very different

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@Harsh, In our counting there is redundancy. Suppose in our first set {1, 2, 3, 4,5 } these element present.
You are selecting first two element suppose you select {1, 2 } . Now after selecting 4 element from other two set. 9 people are left. No restricted is there. So you are selecting 2 people from these 9. Suppose you select 3. 4.
So from the first set you select {1, 2, 3, 4}.

Now think, When you are picking two element, you might pick {3, 4}. Now again select 4 element from other two set. Suppose that these element are same 4 element that you have select for above one. Now again 9 people are left. No restriction. We select from these 9. {1, 2}

Now again from the first set we select {1,2,3, 4}.

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If we consider these three divisions as three objects (variables)and every object is positive (>=0)

X1+X2+X3=8

Than it can be solved as n+r-1Cr

But how to include atleast 2 thing in it

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@bikram sir why after getting number of ways as 6 you took combinations of those? I didn't get that part.

Was that because you were considering employees as distinct, if we consider them indistinguishable from each other in there category then this problem reduce to like three type of donuts and we need have a combination of 8 with atleast 2 each of them, which give us answer as 6 way possible.

But if now we consider each donut having different colour, then how many different colour combinations of such 8 donuts possible with given conditions, then your answer is correct.

Tell me if I have done some mistake.