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## Comment on

Handling Restrictions## Dear Sir

IF the total number of ways in which 10 students can be arranged in a row such that A is always ahead of B?

a. 2x10!

b. 10! /2

c. 10! x 8!

d. none

Regards

## We can arrange 10 students in

We can arrange 10 students in 10! ways.

In HALF of those 10! arrangements, A will be ahead of B, and in the other HALF of those 10! arrangements, B will be ahead of A

So, the number of arrangements where A is ahead of B is 10!/2 (and the number of arrangements where B is ahead of A is 10!/2)

## I didnt understand how in 10!

## We can arrange 10 students in

We can arrange 10 students in 10! ways.

In each arrangement, EITHER A is ahead of B OR B is ahead of A.

Since the there's no reason why one student (A or B) should be ahead of the other student, we can say that, in HALF of the arrangements A will be ahead of B, and in the other HALF of those 10! arrangements, B will be ahead of A.

## As Sir explained that A has

## How will you solve this

A card game using 36 unique cards, four suits(diamonds, hearts, clubs and spades) with cards numbered from 1-9 in each suit. A hand is a collection of 9 cards, which can be sorted however the player chooses. What is the probabilities of getting all four of the 1s?

## Great question.

Great question.

P(getting all four 1's) = (total number of ways to select 9 cards that include all four 1's)/(total number of ways to select 9 cards)

DENOMINATOR first: Total number of ways to select 9 cards.

We can select 9 cards from 36 cards in 36C9 ways

36C9 = 94,143,280

NUMERATOR second: since the four 1's must be in the hand, let's place those four 1's in the hand right away, and then select the remaining 5 cards. Once we place the four 1's in the hand, there are 32 cards remaining. We can select 5 cards from 32 cards in 32C5 ways

32C5 = 201,376

So, P(getting all four 1's) = 201,376/94,143,280

≈ 0.00214

## Can you please explain the

I dont seem to get it.

## Let's use an example:

Let's use an example:

There are 100 people comprised of 60 women and 40 men.

We must choose 1 person from the group to be President.

If the President cannot be a man, then in how many ways can we select a President?

Yes, this is a rudimentary example, but the concept is the same.

(number of VALID outcomes) = (number of outcomes that IGNORE the restriction) - (number of outcomes that BREAK the restriction)

If we IGNORE the restriction, there are 100 ways to choose a person.

We can BREAK the rule/restriction in 40 ways (since there are 40 men)

So, (number of VALID outcomes) = 100 - 40 = 60

Does that help?

Cheers,

Brent

## Thank for you this example!

## Hello. I recently solved the

I approached this drawing out the stages: _ _ _ _. One has to be a biography, so there's 4 options. You need another biography, so there's 3 options. Then for slot 3, you have 2 biographies left plus 6 novels so 8 options. Finally, in the last slot you have 7 books left. However, the resulting number (4 * 3 * 8 * 7) is definitely wrong. Can you further explain why using the restriction and FCP strategy in this case were not correct? Thank you so much!

## Question link: https:/

Question link: https://greprepclub.com/forum/a-reading-list-for-a-humanities-course-con...

The main issue with that approach is that we end up counting some outcomes more than once.

The show this, let A, B, C, and D represent the four biographies, and let e, f, g, h, i and j represent the six novels.

So some possible outcomes are:

AChB

CAhB

BCAh

BACh

etc

Based on your solution, all of these outcomes are considered different even though we end up with the same 4 books each time

In the video at (https://www.greenlighttestprep.com/module/gre-counting/video/788), we learn a strategy for determining whether we can use the Fundamental Counting Principle FCP) to answer a counting question.

We need to ask a question,"Do the outcomes of each stage differ from the outcomes of the other stages?"

In this case the answer is no.

If we select biography A in the FIRST stage, that book gets added to the reading list.

If we select biography A in the SECOND stage, that book gets added to the reading list.

Since the outcomes are the same, we can't use the FCP.

Instead we need to use combinations.

Here's my solution (that uses combinations): https://greprepclub.com/forum/a-reading-list-for-a-humanities-course-con...

Does that help?

## This is very helpful. Thank

## Glad to hear it!

Glad to hear it!

## Hi Brent,

Thank you for the clear explanations, but I'm confused on this one. If there aren't any restrictions on letters repeating, I'm confused as to why stage 1 is 4 instead of 5.

## Good question!

Good question!

When it comes to questions that ask you to ARRANGE a bunch of objects, it's implied that repetitions are not allowed.

So, for example, when we arrange the five letters (J, K, L, M and N), each letter is used only once.

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