# Math Help - New Question

1. ## choose x distinguishable devices from a set

another question concerning the binomial coefficient from my "discrete mathematics and its applications" book:

"there are 8 sorts of distinguishable devices. How many ways are there to choose a dozen devices with at least 3 of sort A and no more than 2 of sort B." solution: 9724

${7-1+7 \choose 7} + {2-1+8 \choose 2} = 1716+36$

a dozen = 12. 12 - 3 ( at least 3 of sort A ) = 9. 9-2 = 7 - so i can choose 7 out of 7 sorts, this is the first part of the sum. the second part of the sum is my second choise of 2 out of 8 distinguishable devices. i sum them up because there is no intersection.

what did i do wrong ?

2. Once again, the fastest way to this answer is by way of generating functions.
Find the coefficient of $x^{12}$ the expansion of $\left( {1 + x + x^2 } \right)\left( {\sum\limits_{k = 3}^{12} {x^k } } \right)\left( {\sum\limits_{k = 0}^9 {x^k } } \right)^6$.

But you seem not to be using generating functions in the class.

So here is tedious way.
$\binom{9+8-1}{9}$ is the number of ways to select a dozen including at least three of type A.

$\binom{6+8-1}{6}$ is the number of ways to select a dozen including at least three of type A and least three of type B.

Thus, $\binom{9+8-1}{6}-\binom{6+8-1}{6}=9724$
is the number of ways to select a dozen including at least three of type A but no more than two of type B.

3. Originally Posted by Plato
Once again, the fastest way to this answer is by way of generating functions.
Find the coefficient of $x^{12}$ the expansion of $\left( {1 + x + x^2 } \right)\left( {\sum\limits_{k = 3}^{12} {x^k } } \right)\left( {\sum\limits_{k = 0}^9 {x^k } } \right)^6$.

But you seem not to be using generating functions in the class.

.

in class we had tasks like: $(x+y)^{25}$ whats the coefficient of $x^{13}$ ?

but .. i am not able to extrapolate in my brain how to use this in this excercise, if it is not to hard, may be you can try to explain?