Difference when selecting items from a set

Question

I saw this question: There are 4 identical pens and 7 identical books. In how many ways can a person select at least one object from this set?

The answer is: (5*8)-1=39. Makes sense, because I can choose between:

Pens:
0
1
2
3
4
5
6
7

Books
0
1
2
3
4
5
6
7
8

And then -1 because we don't want to match 0 and 0.

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.

dreamer954 · Answer

Please share the solution if you know. Thanks in anticipation !!!

Posted from my mobile device

MahmoudFawzy · Answer

Here are my 2 cents ..

To explain, let assume a simpler question and discuss in more details.

Let’s say there are 2 identical pen and 3 identical books.
The ways to select at least one object are: (3*4 - 1 = 11)
(1p+0b) , (0p+1b) , (1p+1b), (2p+0b) , (0p+2b) , (1p+2b) , (2p+1b), (0p+3b) , (2p+2b), (1p+3b) , (2p+3b)

what if the pens are different?
(1p+0b) --> 2 ways (two different pens)
(0p+1b) --> 1 way
(1p+1b) --> 2 ways
(2p+0b) --> 1 ways (both pens are there, order is not important)
(0p+2b) --> 1 ways
(1p+2b) --> 2 ways
(2p+1b) --> 1 ways
(0p+3b) --> 1 ways
(2p+2b) --> 1 ways
(1p+3b) --> 2 ways
(2p+3b) --> 1 ways

sum = 15

what if both pens and different?
(1p+0b) --> 2*1 = 2 ways
(0p+1b) --> 1*3 = 3 way
(1p+1b) --> 2*3 = 6 ways
(2p+0b) --> 1*1 = 1 ways
(0p+2b) --> 1*3 = 3 ways
(1p+2b) --> 2*3 = 6 ways
(2p+1b) --> 1*3 = 3 ways
(0p+3b) --> 1*1 = 1 ways
(2p+2b) --> 1*3 = 3 ways
(1p+3b) --> 2*1 = 2 ways
(2p+3b) --> 1*1 = 1 ways

sum = 31

IanStewart · Answer

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.

ccooley · Answer

You still multiply together the number of ways to select the pen(s), times the number of ways to select the book(s).

The books are easier to deal with, so start there: there are 5 different ways to do it. You could select 0, 1, 2, 3, or 4 of the books, for a total of five options.

For the pens: this is a pretty different problem! The quickest way to solve something like this is to think about it in this way. For each of the 7 different pens, you could either choose it, or not choose it. That's two different options for pen #1 (chosen or not chosen), two options for pen #2 (chosen or not chosen), two options for pen #3 (chosen or not chosen), etc. For example, here are some possibilities for the 7 pens in order:

in, in, in, in, in, in, out
in, out, in, in, in, in, in
out, out, out, out, out, in, in
in, out, in, out, in, out, out
etc.

In total, with two options for what to do with each pen, you have 2*2*2*2*2*2*2 = 2^7 options.

So, the answer overall will be (5)(2^7) - 1.

By the way, you subtract 1 for the same reason you did initially! You don't want to choose 0 pens and 0 books, so you have to remove that option at the end.

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Difference when selecting items from a set

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