Post your question in the Comment section below, and I’ll answer it as fast as humanly possible.

- Video Course
- Video Course Overview
- General GRE Info and Strategies - 7 videos (free)
- Quantitative Comparison - 7 videos (free)
- Arithmetic - 42 videos
- Powers and Roots - 43 videos
- Algebra and Equation Solving - 78 videos
- Word Problems - 54 videos
- Geometry - 48 videos
- Integer Properties - 34 videos
- Statistics - 28 videos
- Counting - 27 videos
- Probability - 25 videos
- Data Interpretation - 24 videos
- Analytical Writing - 9 videos (free)
- Sentence Equivalence - 39 videos (free)
- Text Completion - 51 videos
- Reading Comprehension - 16 videos

- Study Guide
- Your Instructor
- Office Hours
- Extras
- Prices

## Comment on

MISSISSIPPI Rule## I would like to know, how did

How did you do the multiplication?

## Let's do some elimination.

Let's do some elimination.

First, 11! = (11)(10)(9)...(5)(4)(3)(2)(1) and 4! = (4)(3)(2)(1)

So, we have [(11)(10)(9)...(5)(4)(3)(2)(1)]/[(4)(3)(2)(1)(4)(3)(2)(1)(2)(1)]

Cancel out: [(11)(10)(9)...(5)]/[(4)(3)(2)(1)(2)(1)]

The denominator now equals 48, so rewrite it as...

[(11)(10)(9)(8)(7)(6)(5)]/[(6)(8)]

Cancel out to get: (11)(10)(9)(7)(5)

Evaluate: 34,650

## Pls how did you arrive at 20

## 5!/(3!1!1!) = (5)(4)(3)(2)(1)

5!/(3!1!1!) = (5)(4)(3)(2)(1)/(3)(2)(1)(1)(1)

= (5)(4)

= 20

Alternatively,

5!/(3!1!1!) = (5)(4)(3)(2)(1)/(3)(2)(1)(1)(1)

= 120/6

= 20

## Hi,

help on this ques plz.

In how many different ways can 3 fiction books and 3 non-fiction books be arranged in a row of 6 books on a shelf such that the fiction books are not separated, and the non-fiction books are not separated?

can I apply Mississippi rule

Thanks

## We can't use the MISSISSIPPI

We can't use the MISSISSIPPI rule here, because the objects are not identical. That is, the 3 fiction books are all different, and the 3 non-fiction books are all different.

Here's my full solution: https://gmatclub.com/forum/in-how-many-different-ways-can-3-fiction-book...

Cheers,

Brent

## Hi,

Thanks very much for the link.

Iam bit confused because it is not mentioned that the fiction books are different? will it be mentioned if the books are alike

Thanks

## That's a good point!

That's a good point!

To remove ambiguity, the question should state that the 3 books (of each type) are all different.

The linked question is not an official GRE. The official GRE test-makers would be certain to remove all ambiguity from a question like this.

Cheers,

Brent

## Thanks Brent, I hope they do

## Regarding this problem (https

From a box of 10 lightbulbs, you are to remove 4. How many different sets of 4 lightbulbs could you remove?

How are we supposed to know not to use the Mississippi rule? Unless stated otherwise, it would seem like the lightbulbs were not unique objects and therefore we should use the Mississippi rule.

For example, if we were to represent each lightbulb by a letter, it seems like we would have 10 lightbulbs all labeled "B". Like so: BBBBBBBBBB

If we looked for the number of arrangements possible of these non-unique objects using the Miss. rule we would get:

n!/A! = 10!/10! = 1

## Link: https://greprepclub.com

Link: https://greprepclub.com/forum/from-a-box-of-10-lightbulbs-you-are-to-rem...

Great question!

That was my first reaction when I read the question. It would be help avoid any ambiguity if they added 1 word to get "10 DIFFERENT light bulbs"

As you've suggested, if all 10 bulbs are IDENTICAL, then there will be only ONE way to select 4 bulbs.

That said, on test day, you should assume objects are different, unless stated otherwise.

Cheers,

Brent

## Hello dear Brent

If you have 6 identical balls, and you want to distribute these balls among 3 children, how many different ways we can do that in fast technique?

## That would be a very very

That would be a very very tricky GRE question. That said, here's how we can solve this using the Mississippi Rule.

Let's call the three children A, B and C.

I'm going to arrange the letters O, O, O, O, O, O, I, I, and each arrangement will correspond to a different distribution of balls among the three children.

We can think of the O's as representing the balls, and the I's as SEPARATORS that separate the balls that go to each child.

The number of O's that appear before the first separator (the first I) will be the balls that go to child A.

The number of O's that appear after the first separator but before the second separator will be the balls that go to child B.

The number of O's that appear after the second separator will be the balls that go to child C.

For example, the arrangement OOOIOIOO represents 3 balls going to child A, 1 ball going to child B, and 2 balls going to child C.

The arrangement OOIOOIOO represents 2 balls going to child A, 2 balls going to child B, and 2 balls going to child C.

The arrangement OOIIOOOO represents 2 balls going to child A, 0 balls going to child B, and 4 balls going to child C.

The arrangement OOOOOIOI represents 5 balls going to child A, 1 ball going to child B, and 0 balls going to child C.

The arrangement IOOIOOOO represents 0 balls going to child A, 2 balls going to child B, and 4 balls going to child C.

etc.

So now the question becomes " and how many different ways can we arrange the letters in OOOOOOII?"

Applying to Mississippi Rule we get:

The number of outcomes = 8!/(6!2!) = 28

Cheers,

Brent

## Thanks a million, dear Brent.

## Thanks Omer!

Thanks Omer!

I forgot to mention that we can also solve this question by Listing and Counting (and looking for a pattern).

You can see the process in action with a very similar question here: https://www.greenlighttestprep.com/module/gre-counting/video/774

Cheers,

Brent

## Thank you so much, dear Brent

## Hi brent, can you please

## There are two main reasons

There are two main reasons why we can't use the Mississippi rule with that question:

1) The objects (people) must be arranged in a straight line (not a circle) to apply the Mississippi rule.

2) The objects (people) to be arranged must include identical objects.

So for example, if we want to arrange 4 identical B's and 3 identical G's in the straight line, the number of ways to do this = 7!/(4!3!)

## Add a comment