This is our part 1 on tricky GMAT permutation and combination questions and how to solve them efficiently

Let’s start with an example:

**John has to form a management team from 10 candidates. He can choose to include any number of members from the candidates all the way from 0 to 10. If 0 candidates are chosen, John is the sole member in the management team. Each candidate has a unique skill and their inclusion in the management team would change the nature of the management team. How many different management teams can John form?**

**A. 11**

**B. 1024**

**C. 1280**

**D. 1640**

**E. None of the above**

**Solution:**

We may be tempted to say John can form 11 management teams (i.e., with 0 candidates chosen, 1 candidate chosen,.10 candidates chosen).

However, the question throws a twist saying each candidate has a unique skill – i.e., if John chooses only 1 candidate – he can choose 10 types of candidates – and hence 10 different types of management teams.

Hence, the solution we are after is :

Total management teams possible = (# of choosing 0 candidates) + (# of ways of choosing 1 candidate) + (# of ways of choosing 2 candidates) + … + (# of ways of choosing 10 candidates)

Total management teams possible = ^{10}C_{0} + ^{10}C_{1 }+ ^{10}C_{2 }+ ^{10}C_{3 }+ ^{10}C_{4 }+ ^{10}C_{5 }…+ ^{10}C_{10}

Now, ^{n}C_{r} = (n!)/(r!(n-r)!)… we know this from basics of permutations and combinations

Using the above formula to compute ^{10}C_{0} + ^{10}C_{1 }+ ^{10}C_{2 }+ ^{10}C_{3 }+ ^{10}C_{4 }+ ^{10}C_{5 }…+ ^{10}C_{10} is going to be a nightmare J

**Trick : ^{n}C_{0} + ^{n}C_{1 }+ ^{n}C_{2 }+ ^{n}C_{3 }+ ^{n}C_{4 }+ ^{n}C_{5 }…+ ^{n}C_{n }= 2^{n}**

So, the solution is 2^{10} = 1024

We go with **option (B)**

Why does this formula work?

Imagine the 10 candidates standing in a line and you could go from left to right and while passing each candidate – either decide yes or no. Yes means they are included and no means not.

yes yes yes yes yes yes yes no yes yes is a possibility. And so are :

yes yes yes no no yes yes no yes yes

yes no no no no yes yes no yes yes

…. many others….

# of overall possibilities = 2 (can be yes or no) x 2 x 2 ….x 2 (10 times) = 2^{10}

If we generalize this, then for n candidates/items, we are left with :

^{n}C_{0} + ^{n}C_{1 }+ ^{n}C_{2 }+ ^{n}C_{3 }+ ^{n}C_{4 }+ ^{n}C_{5 }…+ ^{n}C_{n }= 2^{n}

This is quite a powerful formula and worth remembering.

Another example of a question :

John is making a cake for his sister’s birthday. He has a box of 7 ingredients – he has to use at least 1 ingredient and can choose as many ingredients as he wants. How many different combinations of ingredients can he use for the cake?

Answer : 2^{7} – 1

Why did we take the 1 out of 2^{7} ? This is because he has to use at-least 1 ingredient so the 1 case with **no no no no no no no** is not possible and has to be subtracted from the total # of possibilities 🙂