While discussing Permutations and Combinations many months back, we worked through several examples of arranging people in seats. Today we bring you an interesting question based on those concepts. It brings to the fore the tricky nature of both Data Sufficiency and Combinatorics – so much so that when the two get together, it is unlimited fun!

In some circumstances, we suggest you to travel the whole nine yards – i.e. solve for the answer in Data Sufficiency questions too even if you feel that sufficiency has already been established. This is especially true for quadratic equations which we assume will give us two values of x but might actually give just a single unique value (such that both roots are the same). In Combinatorics too, sometimes what may look like two distinct cases could actually give the same answer. Let’s jump on to the question.

**Question 1**: There are x children and y chairs in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12

Statement 2: There are more chairs than children.

**Solution:**

There are x children and y chairs.

x and y are prime numbers.

Statement 1: x + y = 12

Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:

Case 1: x=5 and y=7

There are 5 children and 7 chairs.

Case 2: x=7 and y=5

There are 7 children and 5 chairs

At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.

Case 1: x=5 and y=7

If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.

Total number of arrangements would be 7C5 * 5!

Case 2: x = 7 and y = 5

If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.

Total number of arrangements would be 7C5 * 5!

Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.

So actually this statement alone is sufficient! Most people would not have seen that coming!

Statement 2: There are more chairs than people.

We don’t know how many children or chairs there are. This statement alone is not sufficient.

Answer: A

We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!

Now, what if we alter the question slightly and make it:

**Question 2:** There are x children and y chairs *arranged in a circle* in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12

Statement 2: There are more chairs than children.

*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 the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Couldn’t you also have 1 and 11 as candidates for x and y?

You are given that x and y are prime numbers. Since 1 is not prime, we cannot have x as 1.

For question 2:

5 children, 7 chairs – The 1st child can be seated in any chair and the other 4 children can be seated in the remaining 6 chairs in 6*5*4*3 ways.

7 children, 5 chairs – The 1st child can be seated in any chair and the other 6 children can be seated in the remaining 4 chairs in 6*5*4*3 ways.

So,it seems, we have similar counts in circular arrangement as well?

For the circular permutation, it would be (n-1)! ways of arranging n objects.

So, considering cases 1 and 2 similar to above, we would have

7C5 ways of choosing 5 children (or chairs) to match 7 chairs (or children).

Since the arrangement is circular, there are 4! distinct ways of arranging the 5 children in 5 circular seats.

So the answer is 4!*7C5