There are 4 identical pens and 7 identical books. In how many ways can

Hi,
Bunuel, choice D is missing 1 in 2^4-1..

the Q talks of identical things and we should not confuse with Combinations formula for different items..

solution-

7 items can be choosen in 8 ways- 0,1,2..7 at a time
similarily 4 items can be choosen way will be 5- choosing 0, choosing 1.. choosing 4

total ways = 8*5=40..
But we are looking for ATLEAST 1, so subtract 1 way when both are 0..
ans 40-1=39..
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Edited. Thank you.
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Selecting pen 1 is the same as say selecting pen 2 or 3 or 4 because they are identical- So there is just one way of selecting 1 pen
Similarly there is just 1 way each of selecting 2 ,3 or 4 pens
One can also not select any pen. So there are 5 ways of selecting pens.
Similarly there are 8 ways of selecting books.
Total number of ways = 8*5=40.
Selecting at least one object = Total number of ways of selecting - ways of selecting no object
=40-1=39
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Sir, can you please explain 1) why you subtract 1 from side? Is there any general rule? 2) how do you derive 5 ways for pen and 8 ways for books?

Thanks in advance.

Sent from my XT1663 using GMAT Club Forum mobile app

Recall that “at least one” means “one or more.”

Let’s say there are only 1 pen and 2 identical books. The ways to select at least one object are:

1p, 1b ; (1p + 1b), 2b ; (1p + 2b)

We see that there are 5 ways to do this (note: the commas separate the ways to select a certain number of objects and the semicolons separate the number of objects to be selected).

Let’s say there are 2 identical pen and 3 identical books. The ways to select at least one object are:

1p, 1b ; (1p + 1b), 2p, 2b ; (1p + 2b), (2p + 1b), 3b ; (2p + 2b), (1p + 3b) ; (2p + 3b)

We see that there are 11 ways to do this.

Since 5 is 1 less than 6 = 2 x 3 = (1 + 1)(2 + 1) and 11 is 1 less than 12 = 3 x 4 = (2 + 1)(3 + 1). It appears that when there are a identical objects of one sort and b identical objects of another sort, the number of ways to select at least 1 object is (a + 1)(b + 1) - 1.

Therefore, the number of ways to select at least one object from 4 identical pens and 7 identical books is (4 + 1)(7 + 1) - 1 = 40 - 1 = 39.

Answer: C
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Take the task of selecting items break it into stages.

Stage 1: Decide how many identical pens to select
We can chose 0, 1, 2, 3 or 4 pens
So, we can complete stage 1 in 5 ways

Stage 2: Decide how many identical books to select
We can chose 0, 1, 2, 3, 4, 5, 6, or 7 books
So, we can complete stage 2 in 8 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus make our selection) in (5)(8) ways (= 40 ways)
HOWEVER, our above solution allows us to choose 0 pens in stage 1 and 0 books in stage 2.
This 1 outcome is not allowed, because we're told we must select at least one object
Since that 1 outcome is not allowed, the total number of outcomes = 40 - 1 = 39

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Thank you for an interesting question Bunuel !

Any idea how I can solve these (below)?

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?

I replied to this question here:

https://gmatclub.com/forum/difference-w ... 96917.html
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11c2 - 4c1-7c1 = 39

i.e total ways-selecting none of each pair

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