In how many ways can Nylah select 2 staplers, 5 notepads, and 4 boxes of pens from a cabinet with 8 staplers, 10 notepads, and 5 boxes of pens?

Answer 1

if the page is cutting out part of the test highlight

and copy all the text and paste it in an application

like notepad or google docs.

To do this problem one needs to understand the following topics:

1. sets

2. subsets

3. powersets

4. cardinality of a set (aka size of a set)

5. the cartesian product

6. the size of a cartesian product (also known as the product rule)

7. permutations

8. permutation trees

10. naive permutation generating algorithm (optional)

11. permutation formula

12. combinations

13. combination formula

14. naive combination generating algorithm (optional)

15. pascals triangle

16. simplist combination and permutation problems (as building blocks

to solve harder versions of those problems)

Anyways, we need to break down the problem, technically these are 4 sub problems,

3 of the sub problems are identical and are combination problems. The last sub

problem combines the 3 solutions into one big solution.

First analyze that we need to take 2 staplers from 8 staplers, in other

words we want some random subset of size 2 {x_i, x_j} from the

set {x_1,_x2,x_3,x_4,x_5,_x_6, x_7,x_8,x_9}, where x_i and x_j are symbolic

representations for some random pair of staplers. Recall that sets don't have

an ordering, therefore making this problem a combination problem. We can therefore

calculate the number of possible ways we can select 2 staplers from 8 staplers using

the formula C(n,r) = n!/(r!(n - r)!), where n is the size of the set we are grabbing

the items from, and r is the number of items we want to pick. Plugging in numbers

we can see we get

C(8,2) = 8!/(2!(8 - 2)!) = (8 x 7 x 6 x 5 x 4 x 3 x 2 ) /(2!(6 x 5 x 4 x 3 x 2))

= (8 x 7)/2 = 56 / 2 = 28. Therefore the number of possible ways we can select 2

staplers from 8 staplers is 28.

For now its more convenient to represent this in

combination notation, so we will represent that here by doing C(8,2) (which is the

same as 28). Now the same reasoning applies to the notepad and the boxes of pens so

I'll skip straight towards their respective answers. The numbers of ways we can

select 5 notepads from from 10 notepads is C(10,5) and the number of ways we can

select 4 boxes of pens from 5 boxes of pens is C(5,4). So finalizing what we have

we now have three possible collections C(8,2) (staplers), C(10,5) (notepads),

and C(5,4) (pens).

Now C(8,2) means we have 28 different possible ways of collecting

2 staplers from 8 staplers correct? This means there exists a set with those respective

combinations, all possible subsets of size five from the set of 8 staplers.

{{x_1,x_2,x_3,x_4,x_5}, ..... {x_1, x_6, x_3,x_4,x_5}} is the informal representations

of that set of 28 possible combinations (notice its a set containing sets as elements).

Now the same reasoning applies to the notepads and pens. So getting all of our info we

can imagine sets for the combinations C(8,2) (staplers), C(10,5) (notepads),

and C(5,4) (pens).

Because we have sets, we can associate elements from each

set using the cartesian product. Recall that the cartesian product of 3 sets is

A x B x C = {(a,b,c) | a in A and b in B and c in C} and recall that the size of

that cartesian product is |A| x |B| x |C|, where |A| is the size of set A, |B| is

the size of set B, and |C| is the size of set C.

Therefore the product C(8,2) x C(10,5) x C(5,4) is our answer.

Simplify C(8,2) x C(10,5) x C(5,4), we will get the numerical answer: 35280 ways.

THEREFORE the number of ways Nylah can select 2 staplers, 5 notepads, and 4 boxes of

pens from a cabinet with 8 staplers, 10 notepads, and 5 boxes of pens is 35280 ways.