# Using Symmetry in Probability on the GMAT

We know that Combinatorics and Probability are tricky topics. It is easy to misinterpret questions of these topics and get the incorrect answer – which, unfortunately, we often find in the options, giving us a false sense of accomplishment.

In many questions, we need to account for different cases one by one but we don’t really see such questions on the GMAT since we have limited time. Also, we don’t tire of repeating this again and again – GMAT questions are more reasoning based than calculation intensive. Usually, there will be an intellectual method to solve every GMAT question – a method that will help you solve it in seconds.

We have discussed using symmetry in Combinatorics before. It can be used in many questions though most people don’t realize that. In our ongoing endeavor to expose you to intellectual methods, here we present how most people tackle a question and how you can tackle it instead to be in the top 1%ile.

Question: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?

(A) 5
(B) 6
(C) 11
(D) 16
(E) 19

Solution:

Most Common Solution:

What are the permutations of sequence S? They are the different ways in which we can arrange the elements of S. For example, 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc

In how many different ways can we make the sequence? The first element can be chosen in 4 ways – one of 3, 4, 5 and 6. (You are given that 2 cannot be the first element).

The second element can be chosen in 4 ways (2 and the leftover 3 numbers).

The third element can be chosen in 3 ways.

The fourth element can be chosen in 2 ways.

And finally there will be only 1 element left for the last spot.

Number of ways of making set S = 4*4*3*2*1 = 96

In how many of these sets will 5 be in the second spot?

If 5 is reserved for the second spot, there are only 3 ways of filling the first spot (3 or 4 or 6).

The second spot has to be taken by 5.

The third element will be chosen in 3 ways (ignoring 5 and the first spot)

The fourth element can be chosen in 2 ways.

And finally there will be only 1 element left for the last spot.

Number of favorable cases = 3*1*3*2*1 = 18

Required Probability = Favorable Cases/Total Cases = 18/96 = 3/16 = a/b

a+b = 3 + 16 = 19

Intellectual Approach:

Use a bit of logic of symmetry to solve this question without any calculations.

Set S would include all such sequences as 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc – starting with 3, with 4, with 5 or with 6 with equal probability.
By symmetry, note that 1/4th of them will start with 5 – which we need to ignore – so we are left with the rest of the 3/4th sequences.

Now, in these 3/4th sequences which start with either 3 or 4 or 6, 5 could occupy any one of the 4 positions – second, third, fourth or fifth with equal probability. So we need 1/4th of these sequences i.e. only those sequences in which 5 is in the second spot.

Probability that 5 is the second element of the sequence = (3/4)*(1/4) = 3/16

Therefore, a+b = 3+16 = 19