# Find how many different arrangements are possible?

I need some help with probability question.Can you help me with this? Below is my question, question 1(i) I get the answer, just question 1(ii)I can't get the answer.

Michael has 4 printed articles download from different online publishers. They are IEEE xplore, Elsevier, Proquest and Emerald Insight. (i) How many different arrangements are possible if he wishes to rearrange the Printed articles?

Answer: n! = 4!
= 4 x 3 x 2 x 1
= 24


(ii) Assume Proquest is one of the sub unit from Emerald Insight, how many different arrangements are possible?

Answer: ?


Can you show me the solve solutions? Thanks for your helping. Wish you have a nice day!

• "Proquest is one of the sub unit from Emerald Insight" - what does that mean? In any case the answer can't be more than $24$ so you could just list all possibilities. – David Apr 24 '19 at 8:16
• I also confuse this question. – Ason Chee Soon Apr 24 '19 at 8:19
• Maybe it means that we should consider the Proquest publications to be indistinguishable from those of Emerald Insight, in which case we want to divide by the number of permutations of the set of these indistinguishable publications... In this case, divide by $2!$. – 雨が好きな人 Apr 24 '19 at 8:28
• That mean is 4!/2! right? – Ason Chee Soon Apr 24 '19 at 8:31
• Yes. I’m not sure exactly what a ‘sub unit’ is in this case though. – 雨が好きな人 Apr 24 '19 at 8:39

As I said in my comment, I’m not sure what is meant by ‘Proquest is one of the sub unit from Emerald Insight’.

However, if it means that that Proquest is to be considered a sub-publication of Emerald Insight, then maybe we want to consider the possible arrangements where the two publications are the same.

As you said, there are $$4!$$ possible permutations of 4 articles. If we consider two of these articles to be indistinguishable, meaning that we, for example, consider the two permutations (IEEE Explore, Elsevier, Proquest, Emerald Insight) and (IEEExplore, Elsevier, Emerald Insight, Proquest) to be the same permutation, then we want $$\frac{4!}{2!}$$.

Generally, if we have $$n$$ objects and several sets of objects with the number of objects in each set given by $$a_1, a_2, \dots, a_m$$ where the objects in a given set are considered to be indistinguishable from the other objects in the set (but objects from different sets are always considered different), then the number of permutations is given by

$$\frac{n!}{a_1! a_2! \dots a_m!}$$

Intuitively, we are starting with the original permutation where all objects are different, and then dividing by the number of permutations of each set of indistinguishable objects.

In your case, we have $$n = 4$$ and we only have one set of two objects so $$a_1 = 2$$. The number of permutations is then $$\frac{4!}{2!} = 12$$.

• I also don't understanding what is meant by ‘Proquest is one of the sub unit from Emerald Insight’.Anyhow, thanks your given answer. – Ason Chee Soon Apr 26 '19 at 6:19