BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability Help

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Sun Jan 13, 2008 6:54 pm

Probability Help

by kenyan_sun » Mon Feb 11, 2008 5:47 am
Hi,
Can someone please explain to me the 'C' in probability. How do i calculate the number of different ways 5 students and 2 teachers can be arranged for example.
Secondly, is there another letter (i think its P. but i could be wrong) that i need to be aware about.

Thanks in advance.

Cheers
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Mon Feb 11, 2008 4:36 pm
"C" and "P" stand for combinations and permutations, which are different concepts than probability (although sometimes the concepts overlap on specific questions).

We use the combinations formula when we're counting unordered subgroups. If we have n total items and we want to know how many different subgroups of k items we can make, the formula is:

n!/k!(n-k)!

Quite often combinations are expressed with a big C and we'd say:

nCk.

We use the permutations formula when we're counting ordered subgroups. If we have n total distinct items and we want to know how many different ways we can arrange k of them, the formula is:

n!/(n-k)!

Quite often permutations are expressed with a big P and we'd say:

nPk.

In a simple situation where you're arranging n distinct items, the total number of possible permutations is n!.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Thu Feb 07, 2008 4:49 pm
Location: Pittsburgh PA

Is the Answer 10?

by wreelp » Fri Feb 22, 2008 2:51 pm
I wrote out my quick calculation but i dont want to write it out and confuse you unless its correct...
Justin R Miller
Top
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Sun Feb 24, 2008 7:48 am

i think the answer is 21

by shwetonline » Sun Feb 24, 2008 9:49 am
According to my understanding,

We have to find out the permutation of 5 student & 2 teacher taken together. i.e. 7!

But we have to exclude the nternal permutation of 5 students & 2 teachers

i.e.

7!/(5!*2!) = 21

Please correct me if I am wrong.

Thanks,
Shweta
starting GMAT preparation
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: i think the answer is 21

by Stuart@KaplanGMAT » Sun Feb 24, 2008 10:26 am
shwetonline wrote:According to my understanding,

We have to find out the permutation of 5 student & 2 teacher taken together. i.e. 7!

But we have to exclude the nternal permutation of 5 students & 2 teachers

i.e.

7!/(5!*2!) = 21

Please correct me if I am wrong.

Thanks,
Shweta
We actually need more information before we can answer the question.

If we just care about where the "T"s and "S"s go in our arrangement, then your solution is 100% correct.

The generic formula for permutations in which we have non-distinct items is:

n!/r!s!t!...

where n is the total number of objects and r, s and t represent duplicates.

On the GMAT, this issue arises frequently in word jumble questions.

For example:

How many different ways are there to arrange the letters of the word "DESERT"?

n=6 and r=2 (two "E"s), so 6!/2!

How many different ways are there to arrange the letters of the word "DESSERT"?

n=7, r=2 (Es), s=2 (Ss), so 7!/2!2!

How many different ways are there to arrange the letters of the word "DESSERTS"?

n=8, r=2 (Es), s=3 (Ss), so 8!/2!3!

* * *

However, if each teacher and student is distinct and we care about where we put each individual person, the answer would simply be 7!=5040.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Thu Feb 07, 2008 4:49 pm
Location: Pittsburgh PA

Number of ways

by wreelp » Sun Feb 24, 2008 4:53 pm
hello, now unless im missing something, there are 2 events (1. students and 2. teachers)
there are 5 possibilities for event 1 (students) and 2 possibilities for event 2 (teachers). So in order to calculate the number of possibilities for 1 student and 1 teacher, you just need to multiply the possibilities for each event.

so you have 5 x 2 = 10

10 different ways one student and one teacher can be arranged.
Justin R Miller
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Number of ways

by Stuart@KaplanGMAT » Sun Feb 24, 2008 7:43 pm
wreelp wrote:hello, now unless im missing something, there are 2 events (1. students and 2. teachers)
there are 5 possibilities for event 1 (students) and 2 possibilities for event 2 (teachers). So in order to calculate the number of possibilities for 1 student and 1 teacher, you just need to multiply the possibilities for each event.

so you have 5 x 2 = 10

10 different ways one student and one teacher can be arranged.
"Arrange" means we care about order, which means permutations.

You solved a completely different question:

If you have 5 teachers and 2 students, how many different ways can you select 1 of each?
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Thu Apr 10, 2008 1:20 pm
Thanked: 1 times

by humeixia » Fri Apr 11, 2008 3:54 pm
thanks, stuart
Top
Post new topic Post Reply
• Page 1 of 1