pencil, pen, and eraser combination, w/ pens of same colour

[defeated_soldier]

There are 7 pens, 5 pencils, and 4 erasers. Of the 7 pens, 3 are black and 4 are blue. In how many ways can we choose at least 2 pens, at least 2 pencils, and exactly 1 eraser, if the pens chosen are of the same colour?

My solution: The 2 pens, 2 pencils, and 1 eraser could be chosen as below:

. . .7C2 . 5C2 . 4C1

I multiplied, rather than adding, because there is the "and" keyword in the exercise. However , this is not the answer that the book gives. Where did I go wrong ?

I will agree, by the way, that there is a keyword "at least" which I have ignored. Bit I don't know what do do with this...?

Could you please tell me how far I am from the actual answer? Thank you!

 

[stapel]

The following is my take on this exercise, but -- fair warning! -- I'm notoriously bad at these. If another tutor suggests something else, go with that instead of this.

(Note: In what follows, I am assuming that, other than the two colors of pens, the pens, pencils, and erasers are indistinguishable.)

Consider the various cases.

i) If the pens are black, then you can choose 2 or 3 pens. Choose 2. Then figure out how many different sets of pencils (two or more) you can choose. (Since you are choosing only one eraser, and no distiguishing information is given about them, I would assume that you can ignore them in your counts. Whatever sets of pens and pencils you come up with, you'll be chucking an eraser in, and then you're done.)

ii) Note that you can also choose three pens. Adding the three pens to the given number of collections of pencils doubles those sets, so multiply the number of options in (i) by "2".

iii) If the pens are blue, then you can choose 2, 3, or 4 pens. You already have, from (i) above, the number of sets of pencils that can go with a set of pens. You now have three more sets of pens that can be added to the sets of pencils. So multiply the number in (i) by "3".

Add the values from (ii) and (iii) to get the total number of sets.

Eliz.

 

[pka]

\(\displaystyle \L \left[ {4\sum\limits_{k = 2}^5 {\left( {\begin{array}{c}
5 \\
k \\
\end{array}} \right)} } \right]\left[ {\sum\limits_{j = 2}^3 {\left( {\begin{array}{c}
3 \\
j \\
\end{array}} \right)} + \sum\limits_{j = 2}^4 {\left( {\begin{array}{c}
4 \\
j \\
\end{array}} \right)} } \right]\)