# Thread: Help with combinatorics problem

1. ## Help with combinatorics problem

Hi

I would appreciate if someone could explain how to solve the following problem:

"How many different five letter "words" can you write with the letters BILBANA"

The answer is 690 but I'm not entirely sure how to get there. I divide the problem into different cases:

A+B = 5!
A+A = 5!/2!
B+B = 5!/2!
A+B+B = ?
A+A+B = ?
A+A+B+B = ?

2. Hello Drdumbom
Originally Posted by Drdumbom
Hi

I would appreciate if someone could explain how to solve the following problem:

"How many different five letter "words" can you write with the letters BILBANA"

The answer is 690 but I'm not entirely sure how to get there. I divide the problem into different cases:

A+B = 5!
A+A = 5!/2!
B+B = 5!/2!
A+B+B = ?
A+A+B = ?
A+A+B+B = ?

You're right so far. Let me continue:

With A + B + B there will be two other letters chosen from the remaining 3. So we have two processes:

• Choose 2 out of 3 remaining letters. This can be done in ${^3C_2}=3$ ways.

• Arrange the 5 letters, which include a repeated B. This can be done in $\frac{5!}{2!}= 60$ ways.

So A + B + B can be done in $3\times 60 =180$ ways.

Similarly A + A + B can be done in $180$ ways.

And then in a similar way A + A + B + B + one other letter can be chosen and arranged in ${^3C_1}\times\frac{5!}{2!2!}=90$ ways.

Therefore the total no of 'words' $= 120 + 60 + 60 + 180 + 180 + 90 = 690$.