What is the total number of ways of selecting at least one object from 2 sets of 10 identical objects picking?

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.

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? a. 12

b. (24 – 1)(27 -1)

d. 211 – 1

Correct Answer

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.

A combination is a selection of r items from a set of n items such that we don't care about the order of selection.

Examples of Combinations

Combinations without repetitions

Let's say that we wanted to pick 2 balls out of a bag of 3 balls colored red (R), green (G) and purple (P). 123
How many unique combinations will we have if we cannot repeat balls?

3 different ways. Our options are: RG, RP and GP.
12 13 23

We can count the number of combinations without repetition using the nCr formula, where n is 3 and r is 2.

We can see examples of this type of combinations when selecting teams for a sports game or for an assignment. We cannot select a team member more than once (so we can't have a team with Danny, Danny and myself) and we do not care about who is selected first to the team (so if I am in a team with Bob and Tom it is the same to me as being in a team with Tom and Bob).

Combinations with repetitions

Let's say that we wanted to pick 2 balls out of a bag of 3 balls, colored red (R), green (G) and purple (P) 123
If each time we select a ball we place it back in the bag, how many unique combinations will we have?

6 different ways. Our options are: RR, RG, RP, GG, GP and PP.
11 12 13 22 23 33

We can count the number of combinations with repetitions mathematically by using the combinations with repetitions formula where n = 3 and r = 2.

We can see examples of this type of combinations when buying ice cream at an ice cream store since we can select flavors more than once (I could get two, three or even four scoops of chocolate ice cream if I wished) and I don't care about which scoop goes on top (so chocolate on top and vanilla on the bottom is the same to me as vanilla on top with a chocolate base).

Permutations Calculator

What is a permutation?

A permutation is a selection of r items from a set of n items where the order we pick our items matters.

Examples of permutations

Permutations without repetitions

Let's say that we wanted to pick 2 balls out of a bag of 3 balls colored red (R), green (G) and purple (P) 123
How many unique permutations will we have if we cannot repeat balls?

6 different ways. Our options are: RG, GR, RP, PR, GP and PG.
12 21 13 31 23 32

We can show this mathematically using the permutations formula with n = 3 and r = 2

We can see examples of this type in real life in the results of a running race (assuming that two people can't tie for the same place) as we clearly care if we come first and our competitor comes second or if it is the other way around.

Permutations with repetitions

Let's say that we wanted to pick 2 balls out of a bag of 3 balls colored red (R), green (G) and purple (P). 123
If each time we select a ball we place it back in the bag, how many unique permutations will we have?

9 different ways. Our options are: RR, RG, GR, RP, PR, GG, GP, PG and PP.
11 12 21 13 31 22 23 32 33

We can show this mathematically by using the permutations with repetitions formula with n = 3 and r = 2.
# permutations = nr = 32 = 9

We can see this in real life in the number of codes on a safe - we can repeat numbers if we want (and have a password such as 1111) and we care about the order of the numbers (so if 1234 opens the safe, 4321 will not).

Explaining the combinations and permutations formulas

How many ways do we have of ordering n balls?

If we have 3 balls colored red (R), green (G) and purple (P) then there are 6 different ways. We have 3 options for the first color, then 2 options for the second color and one choice for the last color. Therefore we have 3 * 2 * 1 different options or 3! For 4 balls, we have 4! different permutations available. For 5 balls we have 5! different options, etc. For n balls we have n! options.

Explaining the permutations formula

How many permutations are there for selecting 3 balls out of 5 balls without repetitions? We can select any of the 5 balls in the first pick, any of the 4 remaining in the second pick and any of the 3 remaining in the third pick. This is 5 * 4 * 3 which can be written as 5!/2! (which is n! / (n - r)! with n=5, r=3).
There is also an alternative way to pick a selection of 3 balls. Let's say we wanted to pick balls 123. Then we could go on to pick the remaining 2 balls too. This would give us the possible permutations 12345 and 12354. We can see that there are 2! (which is 2) different ways of selecting 5 balls if we want 123 to be the first 3 selections. Therefore, we can obtain then number of selections of 3 balls from 5 balls by dividing 5! (the total number of selections) by 2! (permutations in the list of 5! options which begin with 123, or any other 3 balls you may choose). . How many 5 ball permutations will it start? Well 2! because for this selection you have two balls left and they can be arranged in 2! different ways (as we saw above). Therefore to get the number of permutations of 3 balls selected from 5 balls we have to divide 5! by 2!.

Explaining the combinations formula

Each combination of 3 balls can represent 3! different permutations. Therefore, we can derive the combinations formula from the permutations formula by dividing the number of permutations (5! / 2!) by 3! to obtain 5! / (2! * 3!) = 10 different ways. This generalises to other combinations too and gives us the formula #combinations = n! / ((n - r)! * r!)

Explaining permutations with repetitions formula

If we again picked 3 out of 5 balls but with repetitions then we have 5 options for each selection, giving us 5 * 5 * 5 = 125 selections overall. The general formula is therefore #permutations = nr.

Explaining combinations with repetitions formula

Let's see how many combinations there are for selecting 3 balls out of 5 (red (R), green (G), purple (P), turquoise (T) and yellow (Y)) with repetitions. You will notice that our trick from the normal combinations formula does not work. For example, if we look at the combination of two red balls and one green ball only has 3 possible permutations (RGG, GRG, GGR) instead of 3! = 6, since the green appears twice. Therefore we cannot just divide the number of permutations by 6! and be done. Instead we will use a nice representation to make our task easier. We can represent selections in a table so if we wanted to select 2 reds and a green ball we might note it as: R | G | P | T | YOO | O | | |Which can be written more compactly, by omitting the header and unnecessary spaces, as OO|O|||and selecting one green, one purple and one yellow ball can be written as:R | G | P | T | Y| O | O | | Owhich can be written more compactly as |O|O||OFinally, selecting 3 turquoise balls can be written in a table like this:R | G | P | T | Y| | | | OOOwhich can be written as ||||OOO

Each string of 4 |'s and 3 O's corresponds to a selection and vice versa. Therefore the number of ways of selecting 3 balls out of 5 with repetition and where order matters is the same as the number of ways of writing strings from 4 |'s and 3 O's. To figure out how many of these there are, we can start from 7! and then see that we need to divide by 4! because we repeat strings 4! because of | repetition (since initially we treat the 4 |'s as separate symbols) and divide by 3! since we repeat strings 3! times because of O repetition. Therefore there are 7!/(4!3!) different combinations = (n + r - 1)! / ((n - 1)! * r!), which is the formula that we are after.

Combinations versus permutations, what's the difference?

The difference is whether we care about the order. With combinations, the order does not matter. If we had to pick a sports team then the order in which we pick players does not matter. If we do care about the order then we are choosing a permutation. If instead of a sports team we looked at the results of a running race then order becomes important. We do care if we come first and our main contender comes second or vice versa, even though these would be part of the same combination.

How to use the combinations and permutations calculator?

Order is important: defines whether you want to use the combinations calculator (when it's not active) or the permutations calculator (when it's active).

With repetitions: allows you to select combinations and permutations with repetitions (active) or without (inactive).
This is relevant both the combinations calculator and the permutations calculator.

Identical items: allows you to specify if your problem has some repetitions of items but not infinite replacement (active) or whether it does not (inactive). When it's active, you can fill in the number of repetitions for each item. Note that in this case the number of items textbox will represent the number of unique items.
The identical items switch is relevant both to the combinations calculator and the permutations calculator.