# Math Help - Combinations

1. ## Combinations

I am just seeking confirmation to my answers. Thanks in advance!

Here is the question.

A company employs 8 persons in department A, 5 in department B and 3 in department C. A project team comprising of 6 people is required. How many ways can this team be assembled if:

a) From every department, 2 persons are picked.
b) A minimum of 2 are picked from department A and
c) A representative for every department is picked.

My attempts:

a) $\left(^{8}\mathcal{C}_2\right)\left(^{5}\mathcal{C }_2\right)\left(^{3}\mathcal{C}_2\right)$

b) $\left(^{8}\mathcal{C}_2\right)\left(^{14}\mathcal{ C}_4\right)$

c) $\left(^{16}\mathcal{C}_6\right)$

Am I on the right track or I veered off somewhere along the way?

2. Your answer to part a) is correct.

This is part b) $\sum\limits_{k = 2}^6 {\left( {^8 C_k } \right)\left( {^9 C_{6 - k} }\right)}$

Now you do part c). Your answer is not correct.

3. Originally Posted by Plato
Your answer to part a) is correct.

This is part b) $\sum\limits_{k = 2}^6 {\left( {^8 C_k } \right)\left( {^9 C_{6 - k} }\right)}$

Now you do part c). Your answer is not correct.
I did not understand part b. The first part where it is $\left(^{8}\mathcal{C}_2\right)$ is clear, choosing 2 from department A. The second part is $\left(^{9}\mathcal{C}_4\right)$. Here also I understand the subscript number four, which is the number of representatives remaining since we already chose 2. But why is the superscript 9?

Question c is a bit tricky for me because there is no number given. You are just told picking a representative for every department. It could be picking 3 for one department and 1 for the remaining departments or 2 for each or or ... I thought adding all the numbers (8+5+3) and choosing 6 from the total will be right.

4. Originally Posted by Keep
I did not understand part b. The first part where it is $\left(^{8}\mathcal{C}_2\right)$ is clear, choosing 2 from department A. The second part is $\left(^{9}\mathcal{C}_4\right)$.

Question c is a bit tricky for me because there is no number given.
For part b) we can choose 2, 3, 4, 5 or 6 from department A
There are seventeen employees in all, eight in department A and then nine not in department A.
If we choose only two from department A, $\left(^{8}\mathcal{C}_2\right)$, then we must choose four not in department A, $\left(^{9}\mathcal{C}_4\right)$.
You do that for each of $k=3,4,5,6$, $\left(^{8}\mathcal{C}_k\right)\left(^{9}\mathcal{C }_{6-k}\right)$
Now you add them up.

Part c) is difficult. There are nine combinations you have to calculate.
$\begin{array}{ccccc}
A &\vline & B &\vline & C \\
\hline
4 &\vline & 1 &\vline & 1 \\
3 &\vline & 2 &\vline & 1 \\
3 &\vline & 1 &\vline & 2 \\
2 &\vline & 2 &\vline & 2 \\
2 &\vline & 3 &\vline & 1 \\
2 &\vline & 1 &\vline & 3 \\
1 &\vline & 2 &\vline & 3 \\
1 &\vline & 3 &\vline & 2 \\
1 &\vline & 4 &\vline & 1 \\
\end{array}$

5. Originally Posted by Plato
For part b) we can choose 2, 3, 4, 5 or 6 from department A
There are seventeen employees in all, eight in department A and then nine not in department A.
If we choose only two from department A, $\left(^{8}\mathcal{C}_2\right)$, then we must choose four not in department A, $\left(^{9}\mathcal{C}_4\right)$.
You do that for each of $k=3,4,5,6$, $\left(^{8}\mathcal{C}_k\right)\left(^{9}\mathcal{C }_{6-k}\right)$
Now you add them up.

Part c) is difficult. There are nine combinations you have to calculate.
$\begin{array}{ccccc}
A &\vline & B &\vline & C \\
\hline
4 &\vline & 1 &\vline & 1 \\
3 &\vline & 2 &\vline & 1 \\
3 &\vline & 1 &\vline & 2 \\
2 &\vline & 2 &\vline & 2 \\
2 &\vline & 3 &\vline & 1 \\
2 &\vline & 1 &\vline & 3 \\
1 &\vline & 2 &\vline & 3 \\
1 &\vline & 3 &\vline & 2 \\
1 &\vline & 4 &\vline & 1 \\
\end{array}$

That is really a lot of work for part c. You said we have 17 employees but we have 16. 8 in dept A, 5 in dept B and 3 in dpt. C which is 16 in total. So if we have 8 employees in A, there is another 8, not 9, employees in B and C.

6. Originally Posted by Keep
That is really a lot of work for part c. You said we have 17 employees but we have 16. 8 in dept A, 5 in dept B and 3 in dpt. C which is 16 in total. So if we have 8 employees in A, there is another 8, not 9, employees in B and C.
You are right. I cannot do addition.
So adjust the numbers above. Change 9 to 8.

There is no change in part c).
Yes that is a lot of work.

7. I want to say that the coefficient of $x^6$ in
$[(1+x)^8-1][(1+x)^5-1][(1+x)^3-1]$ gives the answer.

Explanation:

$(1+x)^8-1=\sum{x}^r$...r varies from 1 to 8
hence this expression contains
$C_{1}^8x,C_{2}^8x^2,C_{3}^8x^3,C_{4}^8x^4,....$

similarly for other terms of the product given above.

Now consider the first case of the table given by $Plato$
It has the no. of ways
$C_{4}^8.C_{1}^5.C_{1}^3$
now this case is covered in the product by the multiplication $x^4$ of first term, $x$ of second and third term.

Hence, all the cases in the table are summed up in the coefficient of $x^6$ in the product.

8. Originally Posted by malaygoel
I want to say that the coefficient of $x^6$ in
$[(1+x)^8-1][(1+x)^5-1][(1+x)^3-1]$ gives the answer.

Explanation:

$(1+x)^8-1=\sum{x}^r$...r varies from 1 to 8
hence this expression contains
$C_{1}^8x,C_{2}^8x^2,C_{3}^8x^3,C_{4}^8x^4,....$

similarly for other terms of the product given above.

Now consider the first case of the table given by $Plato$
It has the no. of ways
$C_{4}^8.C_{1}^5.C_{1}^3$
now this case is covered in the product by the multiplication $x^4$ of first term, $x$ of second and term.

Hence, all the cases in the table are summed up in the coefficient of $x^6$ in the product.
Say you want the number ways to form a team if at least 2 people must come from department A and at least one person must come from department B.

Would you then be looking for the coefficient of the $x^{6}$ term in the exapansion of

$\big( (1+x)^{8}-1-8x\big)\big((1+x)^{5}-1\big)\big(1+x\big)^{3}$ ?

9. Originally Posted by Random Variable
Say you want the number ways to form a team if at least 2 people must come from department A and at least one person must come from department B.

Would you then be looking for the coefficient of the $x^{6}$ term in the exapansion of

$\big( (1+x)^{8}-1-8x\big)\big((1+x)^{5}-1\big)\big(1+x\big)^{3}$ ?
Yes...that would work. That is the real advantage here. You have to form the product only and it will take care of all the cases.