probability s1

why for this combinations question is 3! not used because the order obviously matters because it is a code

the answer is 4 * 3 *2

this is for question 2 i
http://mei.org.uk/files/papers/s108ja_leqa.pdf

why was 3! used for 4C3 but not for 4 * 3*2
i don't really understand combinations, when can they be used?

Reply 1

Kevin De Bruyne

21

Original post by esmeralda123

why for this combinations question is 3! not used because the order obviously matters because it is a code

the answer is 4 * 3 *2

this is for question 2 i
http://mei.org.uk/files/papers/s108ja_leqa.pdf

why was 3! used for 4C3 but not for 4 * 3*2
i don't really understand combinations, when can they be used?

Combinations do suck.

There are 4 ways of choosing the first letter, 3 the second and 2 the third (without replacement, as the letters have to be distinct) hence its 4*3*2.

That's not particularly convincing, so it may be worth revisiting the textbook or finding something online.

Original post by esmeralda123

why for this combinations question is 3! not used because the order obviously matters because it is a code

the answer is 4 * 3 *2

this is for question 2 i
http://mei.org.uk/files/papers/s108ja_leqa.pdf

why was 3! used for 4C3 but not for 4 * 3*2
i don't really understand combinations, when can they be used?

For part (i) there can be no repetitions. So for the first letter you have 4 options, then for second letter you have 3 options, and for last letter you have 2 options. Thus by the product rule for counting, there are

4 \cdot 3 \cdot 2 = 24

possibilities. Alternatively, this is given by the function

4\mathrm{P}3 = \frac{4!}{(4-3)!}

It cannot be 3! because this number simply represents the number of ways you can arrange 3 letters. It doesn't say anything about the fact that you can choose from 4 objects.

(edited 6 years ago)

Reply 3

debbie394

OP

11

Original post by RDKGames

Doesn't say that the order matters, so it doesn't.

For part (i) there can be no repetitions. So for the first letter you have 4 options, then for second letter you have 3 options, and for last letter you have 2 options. Thus by the product rule for counting, there are

4 \cdot 3 \cdot 2 = 24

possibilities. Alternatively, this is given by the function

4\mathrm{P}3 = \frac{4!}{(4-3)!}

It cannot be 3! because this number simply represents the number of ways you can arrange 3 letters. It doesn't say anything about the fact that you can choose from 4 objects.

but aren't permutations used when order matters?

Reply 4

the bear

20

they should have made the question clearer by including

"for instance the codes AGC and CAG are distinct"

:mad:

Original post by esmeralda123

but aren't permutations used when order matters?

Yes. Got distracted and didnt mean to say that it doesnt matter. So it does, hence you can use permutation function.

But yeah it would’ve been nice if they made this clearer

Reply 6

debbie394

OP

11

Original post by RDKGames

Yes. Got distracted and didnt mean to say that it doesnt matter. So it does, hence you can use permutation function.

But yeah it would’ve been nice if they made this clearer