Problem Solving

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

"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!.

I wrote out my quick calculation but i dont want to write it out and confuse you unless its correct...

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

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.

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.

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?

thanks, stuart