An interesting GMAT problem solving practice question in permutation combination. The concept covered in this question is selecting one or more objects from a set of object, all of which are not distinct. The objects not being distinct is what makes this question an interesting one. This is another question that you can use as a template to solve questions that comprise elements that are not distinct.
There are 4 identical pens and 7 identical books. In how many ways can a person select at least one object from this set?
b. (24 – 1)(27 -1)
d. 211 – 1
Choice E. 39 ways.
Demystifying number of ways of selecting when we are presented with identical objects
The 4 pens stated in the question are identical. Let us understand how that makes a difference.
Had these pens been distinct, the number of ways of selecting one pen out of these 4 would have been 4c1 or 4 ways. For e.g., if we name the pens A, B, C, and D – selecting A is different from selecting B.
But because these pens are identical, the number of ways of selecting one pen out of the 4 is just 1 way. It does not matter which one you picked it will appear the same as picking one of the others.
Let us extend the reasoning to selecting two pens. Had these pens been distinct, the number of ways of selecting two pens out of 4 is 4c2 = 6 ways. For e.g., if we name the 4 distinct pens A, B, C, and D – selecting AB is different from selecting BD.
However, because these pens are identical, the number of ways of selecting two pens out of four is also just 1 ways. You take any two out of the four, it is going to appear the same.
What do we have in this question?
4 identical pens and 7 identical books. We have to pick at least one object.
A person can select none or up to 4 identical pens in 5 ways (0 or 1 or 2 or 3 or 4 pens).
A person can select none or up to 7 identical books in 8 ways (0 or 1 or 2 or .. 7 books).
So, one has 5 ways of selecting pens and 8 ways of selecting books. (These include the option of not selecting any pen and any book)
So, a person can select none or all of the objects in 5 * 8 = 40 ways.
However, one of these 40 ways includes the scenario that neither a pen nor book would have been selected. We need to select at least one object. So, let us eliminate that one possibility.
Therefore, number of ways of selecting at least one object from 4 identical pens and 7 identical books = 40 – 1 = 39.
Note: Had the objects been distinct, we could select none or all objects in 211 ways.
Why so? Because each of the objects has 2 choices – being selected or not being selected.
11 objects have 211 outcomes.
However, if we have to select at least one object, we have to eliminate the only outcome in which we select none of the objects.
Had the objects been distinct, the answer would have been 211 – 1.