How Do You Calculate Permutations and Combinations from the Word WINDOWS?

[fork]

If 4 letters are selected from the 7 letters of the word "WINDOWS", calculate the number of possible
a)combinations.
b)arrangemants.
I have got the answer but I still don't understand how to calculate it.
Thanks.

 

[Tide]

What have you tried so far? How would you handle the problem if all the letters in the word were different from each other?

 

[fork]

I know how to do it now.
a)5C4+5C3+5C2
b)5C4 x4!+5C3 x4!+5C2 x4!/2!

right?
Thanks.

 

[forevergone]

I know how to do it now.
a)5C4+5C3+5C2
b)5C4 x4!+5C3 x4!+5C2 x4!/2!
right?
Thanks.

a) is not right. Because you have a sequence of length 4 from 7 different symbols, it should be 7C4.

b) for arrangements, it's a bit different. I was taught to do it the following way:

find the # of repeating letters so it's only really counted once and divide with the overall # of symbols possible so:

7!/2! would be the number of total arrangements because the 7 is the number of letters divided by the 2! which is the repeating the letter, W.

 

[Solidmozza]

Im also having trouble with this type of question. Please help + explain how you got the correct answer anybody.
~I'm just stumped how it says you must pick 4 letters, instead of just using the ones you have to make an arrangement, and how the letters repeat. If it was only just one of these restrictions, eg How many ways can WINDOWS be arranged in a straight line = 7!/2! (Which is what forevergone was incorrectly referring to: I don't think his answer is right because he is not including the restriction of choosing only 4 letters), then I can do all of them, its just both of those restrictions I am confused.

Thanks a lot!

 

[forevergone]

Wait, you're right. Sorry about that. I was considering for all different arrangements of length 7. If you want only of length 4, then it's simply a permutation question. You have 7 symbols from a sequence of length 4 to choose from.

**It should be 7P4

 

[Solidmozza]

But forevergone, there are 2 W's, and so some of the permutations will repeat - eg. WINW and WINW.. how do you know which W is which?

 

[forevergone]

See but that's the thing. There are 2W's but theyre included inside the word, so what permutation does is that it looks like the same words, but the placeholders in which it holds the W is different. Think of a W in the first letter as an apple and the other as a pear. If you switch the order, it'd be:

apple I N pear

If you switched the other W's around:

pear I N apple

It's sort of that general idea. It's because that there are 2 W's in the word that such cases like this arise. What 7P4 does is out of those 4 symbols, it assures that it selects no more that 1 letter once to use in those 4 sequences. No such letter will be repeated again.

 

[preet]

I don't really get what that means... could you please explain your post further, forevergone? TiA

 

[maverick6664]

I know how to do it now.
a)5C4+5C3+5C2
b)5C4 x4!+5C3 x4!+5C2 x4!/2!
right?
Thanks.

Yeah, these are right. Explanations are:

a)
no "w" 5c4 (choose 4 letters among 5 letters other than W)
one "W" 5C3 (1 "W" and choose 3 letters among 5 letters)
two "W"s 5C2

b)
no "w" 5C4*4!
one "W" 5C3*4!
two "W"s 5C2*4!/2! (divide by 2! because there are two "w"s)