Probability of Four Letter Word from set $\{A, B, C, D, E\}$ with no two consecutive equal letters.

My approach to solve this problem is like :

No. of Four letter word from the set is : $5^4$

No. of Four letter word with two consecutive equal letters : $5^4-[5*3*5^2]$

$5*3*5^2$ = (for each of the five letters $*$ there are three positions $*$ with $5^2$ words)

Required Probability = $\frac{5^4-(5*3*5^2)}{5^4}$

Is this the correct answer ? If not please explain why and whether there any better approach to solve this ?

Problem with your approach is that you are double counting.

eg- cases like "AABB" is being subtracted twice in your approach.

(First in A A_ _ and then again in _ _ B B)

Here is a simple way to do this

There are 4 places _ _ _ _

Now choose a letter for first place = 5 possibilities

Now choose a letter for second place = 4 possibilities(as you can't have a letter which was in first place)

Now choose a letter for third place = 4 possibilities(as you can't have a letter which was in second place)

Now choose a letter for fourth place = 4 possibilities(as you can't have a letter which was in third place)

So total combinations = 5*4*4*4

Probability = $\frac{5*4^3}{5^4}$ = $(\frac{4}{5})^3$

• Can you please explain, what my approach is calculating ? or what's wrong with that? Nov 2, 2017 at 6:43
• @user292174 updated. Check now Nov 2, 2017 at 7:06

first letter can be any of the 5 letters. Second letter can be any of the 4 letters that arent the first letter. Third letter can be any of 4 letters that arent the second letter. fourth letter can be any of the 4 letters that arent the third letter.

$\frac{5\cdot 4\cdot 4\cdot 4}{5^4}$

• Can you please explain, what my approach is calculating ? or what's wrong with that? Nov 2, 2017 at 6:49

First letter can be any of given 5 letters. And second letter can be any of four .Third space can have 4 letters except the letter in second position .similarly the fourth letter Then the result will be (5×4×4×4)/5^4