Now that we have discussed both permutations and combinations independently, it’s time to look at questions that involve both. Mind you, these questions are not difficult -– they just involve both concepts. The first one is a circular arrangement question with a tiny twist. The second one requires us to make some cases. It takes a fair bit of patience to work out one case at a time and I doubt that GMAT will give you such a question since it is a little bit of a bore. (Actual GMAT questions have more entertainment value for the test maker and the test taker. They make you think and are FUN to solve) That said, it is a great question to bind together everything that we have learned till now and strengthen your understanding. Let’s start.

Question 1: Seven women and four men have to sit around a circular table so that no two men are together. In how many different ways can this be done?

Solution: Try and think about it for a while. We did a very similar question while working on circular arrangements. In that question, number of women and number of men were equal so we just had to place them in alternate positions. Here, we have fewer men. What do we do now?

Two men cannot sit together but some women will sit together since there aren’t enough men. So, let’s make the 7 women sit around the round table in (7-1)! = 6! ways (We covered the (n-1)! concept in the post on circular arrangements)

Now, how many places do we have for the men? A man can sit between any two women sitting next to each other. How many such pairs of women are there? Since there are 7 women, we have 7 such pairs and hence 7 possible spaces for men. There are two different approaches you can take from here:

Approach 1:

We have 4 men but 7 possible spaces for them. For the first man, we can select a space in 7 ways. For the second man, we can select a space in 6 ways. For the third one, in 5 ways and for the fourth one in 4 ways. So we can arrange the men in 7*6*5*4 ways. This is just our basic counting principle in action.

Approach 2:

Some people like to split up the task into two steps – make the selection, then arrange. Out of 7 spaces, we need to select any 4 for the 4 men. How do you select 4 out of 7? Using basic counting principle and un-arranging concept, we can do it in 7*6*5*4/4! ways (or we can use the formula ^{7}C_{4}). We have selected 4 spaces so now we just want to arrange the 4 men in the 4 spaces. We can do this in 4! ways.

It doesn’t matter which approach you use. The first one uses just the basic counting principle. The second one is used more often by people who are very comfortable with the combinations formula.

The total number of arrangements we get = 6! * 7*6*5*4 or we can write this as 6!*7!/3! to make it a little compact.

Let’s look at the second question now.

Question 2: How many words of 4 letters can be formed from the word “INFINITY”? (They may or may not be actual words in the English language.)

The word INFINITY has 5 distinct letters – I, N, F, T, Y

Repetitions – I, I, I, N, N

The question doesn’t say that all letters of the words have to be distinct. So, you can make a word using all three Is and another letter or two Ns and two Is etc. So you cannot just select any four letters and arrange them. The number of arrangements will vary depending on whether the letters are all distinct or have some repetitions. Let’s look at all possible cases:

Case 1: All letters are distinct (Form: abcd)

From the 5 distinct letters, we can select any 4 (or drop any 1 letter) in 5 ways (you can also use ^{5}C_{4} or 5*4*3*2/4! to arrive at the figure of 5)

We can arrange these 4 selected letters in 4! ways.

Number of ways in which you can make a 4 letter word with all distinct letters = 5*4! = 120 ways

Case 2: Two letters same, others different (Form: aabc)

Only I and N are repeated so we have to select one of them and we have to select 2 of the other 4 (F, T, Y and whatever is not selected out of N and I) letters.

Select one of N and I in 2 ways. Then to choose 2 other letters, pick two from the other 4 letters in 4*3/2 ways (or 4C_{2}) = 6 ways.

Now we have 3 letters and one of them is repeated so in all we have 4 letters. We can arrange 4 letters (with a repetition) in 4!/2! ways (we divide by 2! because one letter is repeated).

Number of ways in which you can make a 4 letter word with one repetition = 2 * 6 * 4!/2! = 144 ways

Case 3: 2 letters, both repeated (Form: aabb)

We have only two letters that are repeated, N and I. We will need to select both of them so the selection can be done in only 1 way.

Since both the letters are repeated, the 4 letter word can be formed in 4!/(2!*2!) = 6 ways

Case 4: 3 letters same, fourth different (Form: aaab)

Only I appears 3 times so it must be selected. We have to select one letter from the other four. We can choose the fourth letter in 4 ways.

Since I is repeated 3 times, the four letters can be arranged in 4!/3! = 4 ways.

Number of ways in which you can make a 4 letter word 4 * 4 = 16 ways

All four letters cannot be the same since no letter appears four times.

Total number of 4 letter words that can be formed using the letters of the word ‘INFINITY’ are 120 + 144 + 6 + 16 = 286 words

The solution is long but very methodical. If you go one step at a time, it is not complicated at all. I will see you next week with some tricky questions. Till then, keep practicing!

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches **GMAT prep** for Veritas Prep and regularly participates in content development projects such as this blog!*

Karishma,

Gj for the post :) it’s great help.

Regarding Question 2.

Shouldn’t we include the possibility

1. N N I {T,F,Y) —> 3*4!/2!

and

2. I I N {T,F,Y} —> 3*4!/2!

Which account to 24*3=72 more arrangements?

Actually, that is again of the form aabc so we don’t need different cases for that. In case 2 itself, it should be ‘select 1 of N and I and select 2 of the leftover 4 (T, F, Y and whatever remains out of N and I)’. I missed adding this so let me edit the post. Thanks for pointing it out. Good job!

Seven women and four men have to sit around a circular table so that no two men are together. In how many different ways can this be done?

First let’s make the 7 women sit around the round table in (7-1)! = 6! ways.So 7 seats are taken and we are left with only 4 seats for the 4 men.

In your explanation you have mentioned that there are 7 places for men.Please explain how to calculate the number of seats available for men after the seats are taken by the women.

Don’t think of seats as chairs kept around the table. Think that the area around the table is empty. When someone comes to sit, he/she brings a chair along. When the first woman comes, she sits anywhere around the table. Every woman creates a seat till 7 of them are sitting. Now between every two women, there is one distinct spot. So there are 7 distinct spots for the 4 men (the men cannot sit together so there is only one spot between every two women). Try to sketch 2-3 different cases to understand what the final arrangement looks like in those cases.

Hi Karishma,

I tried to arrange the 4 men first and the approach became very complicated. Please let me know whether in such questions it is always better to arrange the larger group first.

thanks!!

No, it is not about the larger/smaller group. It is about the constraint you have.The men should not sit together. The women can sit together. If you make men sit first, how will you decide where to make them sit? With one space between them, two spaces etc – it will be very complicated. Instead make women sit first. Now there will be at least one woman between any two men. Your condition is satisfied.

Also, I am not able to understand approach 2 for question number 1. The 7 spaces where men can sit were created after seating the 7 women. We select any 4 of these for seating the 4 men. After that why are we arranging 4 women in 4 spaces again?

I am sorry if my questions come as silly. Just not very good with P n C :(.

thanks in advance for your help!!

In approach 2, we are first selecting 4 spots out of 7 in 7C4 ways and then arranging the MEN in the four spots (women was a typo).

I say that one man will sit to the right of Jasmin, one to the right of Martha, one to the right of Jia and one to the right of Nikki. This is a selection of 4 spots and can be done in 7C4 ways. Now out of A, B, C and D (four men), who will sit to the right of Jasmin? Here, we arrange the 4 men in 4! ways on the 4 spots.

Hi Karishma,

I am little bit confused with case 2. Let me know if this is the correct reasoning (below). I would also appreciate it if you could post alternative approaches to it.

First Select:

We have four spaces, two of which have to be identical letters, so we can select all four in [(2)(1)(4)(3)]/2 = 12 <– 2 options N or I for the first space/letter, 1 for the second since it needs to be identical, 4 for the following one because 4 distinct letters remain and (3) for the last one because 3 distinct letters remain. We divide by two to 'un-arrange' the distinct letters. No need to 'un-arrange' the identical ones, correct?

AND we need to arrange the four letters in four positions dividing it by 2 because two of the letters are the same: [(4)(3)(2)(1)]/2 = 12

So total arrangements: 12*12 = 144

Yes, that is correct, Vanessa.

Hi Karishma,

I didn’t get the part in the second question where you have mentioned that we can not perfom normal combinations ie. we can’t simply select 4 letters out of 8 letters. After all it’s not mentioned distinct or not distinct. So why won’t a simple combination work in this case ? Please help me in understanding this thing.

Depending on which letters you pick, the number of arrangements will be different.

If you pick 4 distinct letters, you can arrange them in 120 ways.

If you pick 2 same and 2 distinct, you can arrange them in 144 ways.

and so on…

If you just pick 4 letters at random without knowing which letters, how will you decide on the number of arrangements?