1. ## combination's of letters

say you have to find how many 4 letter combination's there are of the word "outlook"

how do you do that?
If it where a word with no repeated letters it wouldnt be bad ie 7C4 (Pretty sure) = 7!/(4!3!) = 35

but what is this, maybe it's 6c4 = 6!/(2!2!) = 180? I know that is wrong the answer is 15.

Thanks.

2. Originally Posted by brentwoodbc

maybe it's 6c4 = 6!/(2!2!) = 180? I know that is wrong the answer is 15.
Hi, maybe you are right, 6c4 =15

3. I made a mistake it should have been 6!/2!4! = 15

but when I do it like that for the others it doesnt work, and it wouldnt work if there where two different letters that where repeated.

osborne

7 letter's

so

7!/ (4!(7-4)!) = 35 {thats with no repeated letters} right answer is 25

?

4. Originally Posted by brentwoodbc
"outlook"
Assume no "o" is used= ${4 \choose 4}=1$
Assume 1 "o" is used= ${4 \choose 3}=4$
Assume 2 "o" is used= ${4 \choose 2}=6$
Assume 3 "o" is used= ${4 \choose 1}=4$

This will consider all cases and total =15

5. Originally Posted by brentwoodbc
"osborne"
Assume no "o" is used= ${5 \choose 4}=5$
Assume 1 "o" is used= ${5 \choose 3}=10$
Assume 2 "o" is used= ${5 \choose 2}=10$

All cases= 25