Question

Mahesh has 4 red pens, 3 blue pens, 2 green pens and 2 black pens. how many different arrangements of these pens are possible, if all the pens of same colour must be placed together

Answer 1

Red pens = 4

Blue pens = 3

Green pens = 2

Black pens = 2

Since we have to keep the same colours togetherly so we can consider it as a block so it'll have 4! ways to be arranged in.

But again each of the blocks can have different combinations in them so each of the them will be having n! factorial ways.

If all the pens of same colour must be placed together, then no of arrangements possible = 4! * 4! * 3! * 2! * 2! = 13824.

Answer 2

Let us see the question carefully first so, in together two cases are possible,

Case 1)

If the pens of the same color are identical to each other

Then , the number of arrangements possible are

= 4! × 4! ×3! × 2! × 2!

=13824 ways

Case 2)

When the pens of same color are similar to each other,

Then

We keep each of the same color pens together and consider them as a single entity

So,

Number of arrangements possible are

= 4!

=24 ways

Note: I hope you understand the ambiguity of the question

Just like all oranges are same in the color but you can still determine and differentiate between the oranges when you buy

In general it should be specifically mentioned in the question but here since it is not mentioned i would prefer you to go for case 2)