peggy10 wrote:
But how about if the question was:
1. 7 different pens, 4 identical books. In how many ways can a person select at least one object from this set?
2. 7 different pens, 4 different books. In how many ways can a person select at least one object from this set?
Would be good to know to clear out exactly how to do in different situations.
These aren't really the kinds of counting questions the GMAT asks, but maybe there's something useful to learn from their solutions.
Let's take the simpler question first: if you have 7 different pens, in how many ways can you select one or more pen? Here, we're not putting the pens in any order when we pick them. There are two ways to answer the question, a long way and a very short way:
- we could pick exactly one pen, which we can do in 7C1 = 7 ways
- we could pick exactly two pens, which we can do in 7C2 = (7)(6)/2! ways
- we could pick exactly three pens, which we can do in 7C3 = (7)(6)(5)/3! ways
and so on, so the answer will just be 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7.
But if we look at the problem differently, we can get the answer almost instantly. For the first pen, we have 2 choices, take it or leave it. Same for each other pen. So in total we have 2^7 choices, but we can't count the one situation where we don't take a single pen, so the answer is (2^7) - 1.
Because those two solutions give the same answer, the above is one way to prove this relationship:
7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7 = 2^7
where you could replace the 7 with any other number, adjusting the number of terms on the left side accordingly. If you've ever seen something known as 'Pascal's Triangle', it demonstrates this relationship as well (each row of that triangle sums to a power of 2, and each row contains these various nCk numbers).
So to answer your question 2 first, if all the items are different, it doesn't matter if they're pens or books or penguins or whatever, we have 11 different items in total, and thus (2^11) - 1 ways to choose one or more of them.
For the first question, we have 2^7 ways to choose any number of pens (including zero), and 5 ways to choose any number of books (including zero), so (2^7)(5) ways to choose two items (including zero of each), and thus (2^7)(5) - 1 ways if we must choose at least one thing.
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