In how many ways can a person post 5 letters in 4 letter box

There are 4 places (ABCD) that need to be filled.
Letter 1 can be posted in either A, B, C, or D i.e. 4 ways
Letter 2 can be posted in either A, B, C, or D i.e. 4 ways
Letter 3 can be posted in either A, B, C, or D i.e. 4 ways
Letter 4 can be posted in either A, B, C, or D i.e. 4 ways
Letter 5 can be posted in either A, B, C, or D i.e. 4 ways

So the total no of ways in which 5 letters can be posted in 4 boxes are = 4 x 4 x 4 x 4 x 4 = 4 ^ 5

Thus the answer is D

Hope it helps.

Something is up with this question. Options (B) and (D) are exactly the same.

In my opinion, the correct solution for this is:

Arrange the 5 items in 4 different "baskets":
xxxxx|||

Total of 8!/(5!*3!) = 8.7 = 56

Now we can order these 5 items among themselves: 5! = 120 ways

Answer = 56*120 = 6720 IMO

Thanks Fameatop (1+)
After reading the problem again and realizing what the question was asking, your logic makes perfect sense.

Is caioguima's approach incorrect because we are counting certain letter box combos multiple times when we multiply by 5!?
Would caioguima's method be correct if the order of the letters in the boxes mattered?

Thanks to vips and debayan as well

Fameatop got it right.

For each letter the person has 4 options.
Thus 4*4*4*4*4=4^5=1024

Ans b and d are both correct. (Dude don't mind but watch for the source)

fameatop-You're correct...!Honestly I was having confusion with this...but it appears really easy

Vips0000-You got it too...!
And the source-it's one of the questions in IIM CAT materials of a CAT prep companies...unfortunately it's due to lack of proof-reading you know !

Not much we can do dude but just to point it out...

Take the task of distributing the 5 letters and break it into stages.

Stage 1: Select a box for the 1st letter to go into.
There are 4 available boxes, so we can complete stage 1 in 4 ways

Stage 2: Select a box for the 2nd letter to go into.
There are 4 available boxes, so we can complete stage 2 in 4 ways

Stage 3: Select a box for the 3rd letter to go into.
There are 4 available boxes, so we can complete stage 3 in 4 ways

Stage 4: Select a box for the 4th letter to go into.
There are 4 available boxes, so we can complete stage 4 in 4 ways

Stage 5: Select a box for the 5th letter to go into.
There are 4 available boxes, so we can complete stage 5 in 4 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 letters) in (4)(4)(4)(4)(4) ways (= 4⁵ ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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I need help in my thought process:
why dont be solve it by slot method

- - - - for four letter box. and then , 1st place has 5 letters tro choose from then 2nd letter box had 4 letters to choose from?
5*4*3*2 ?
Please help me clear this doubt.

Each letter has 4 options.
So, total no of ways = 4 x 4 x 4 x 4 x 4 = 4^5

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The rule of arrangements say that the number of ways of arranging/distributing n distinct things into r different things is given as \(r^n\).

Here letters go into the post boxes. Therefore post boxes = n = 4 and letters = r = 5

\(r^n\) = \(5^4\)

Option D


The simplest way in these type of questions is think, who is going into what. Are letters going into the post boxes or post boxes going into the letters.

Here it becomes clear that letter will go into the post boxes.

So each letter can go into any of the 4 post boxes and therefore letter 1, 2, 3 4 and 5 each have 4 options = 5 * 5 * 5 * 5 = \(5^4\).



Take another example of 4 friends who want to stay in 5 hotels. is the answer 4545 or 5454.


Lets understands this. Will the friends go to the hotels or the hotels go to the friends. It would be friends going to the hotels.

So the first friend can stay in any of the 5 hotels, the 2nd too can stay in any 5 and so can the 3rd and 4th person.

Therefore the answer is 5∗5∗5∗5=\(5^4\)

Hope this helps

Arun Kumar
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