Lesson: Handling Restrictions

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

greenlight-admin's picture

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!/2 arrangement A will be ahead of B, please explain?
greenlight-admin's picture

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 to be ahead of B so it does not mean that they have to be together too. A can be two ahead of B too . Similarly three steps. So if there are 10! factorial ways in total , then there will be definitely half of the ways when A is ahead and Half of ways when B is ahead.

How will you solve this problem in the context of unique arrangements?

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?
greenlight-admin's picture

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 'breaking the restriction' part?
I dont seem to get it.
greenlight-admin's picture

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?


Thank for you this example! It really helped clear up the confusion I was having :)

Hello. I recently solved the ETS Official Guide Test 1 and I was confused on Question 11: "A reading list for a humanities course consists of 10 books, of which 4 are biographies and the rest are novels. Each student is required to read a selection of 4 books from the list, including 2 or more biographies. How many selections of 4 books satisfy the requirements?"

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!
greenlight-admin's picture

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:

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 you so much! Your course has improved my GRE score greatly.
greenlight-admin's picture

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.

greenlight-admin's picture

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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