In a box, there are 4 ballpoint pens and 3 fountain pens. How many possible selections can be formed which have at least 2 items of every type of pen?
A. 26
B. 44
C. 52
D. 78
E. 130
Answer: B
Source: e-GMAT
In a box, there are 4 ballpoint pens and 3 fountain pens. How many possible selections can be formed which have at least
This topic has expert replies
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
I find it difficult sometimes to guess the intentions of questions written this way, because "two items of every type of pen" does not make sense, particularly with only two types of pen. But I gather we just want to count how many ways to pick two pens of each type, then two ballpoints and three fountains, then three ballpoints and two fountains, and so on, until we cover every situation where we pick at least two ballpoints and at least two fountain pens. I suppose if someone had memorized a formula for this type of scenario, they could skip the first two lines below, but it's a formula you'd never need on a real GMAT problem, so I'll show how it's derived:
4C2*3C2 + 4C2*3C3 + 4C3*3C2 + 4C3*3C3 + 4C4*3C2 + 4C4*3C3
= 4C2(3C2 + 3C3) + 4C3(3C2 + 3C3) + 4C4(3C2 + 3C3)
= (4C2 + 4C3 + 4C4)(3C2 + 3C3)
= 11*4
= 44
4C2*3C2 + 4C2*3C3 + 4C3*3C2 + 4C3*3C3 + 4C4*3C2 + 4C4*3C3
= 4C2(3C2 + 3C3) + 4C3(3C2 + 3C3) + 4C4(3C2 + 3C3)
= (4C2 + 4C3 + 4C4)(3C2 + 3C3)
= 11*4
= 44
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com