Hi Jon, thanks for dropping by! You’re more than welcome – there’s so much beauty in math, programming and other topics that is often buried under dense proofs. I’m glad you like the site, I want it to be an open forum for learning 
Great posts… but I have another question:
How many 4-letter combinations are there of the letters in each word? a) ONOWAY b) OSBORNE c) OUTLOOK
I’ve been fighting with this for about 3 hours now. The answers in the text are a) 11 b) 25 c)15
I can’t figure out how to manipulate the formula to account for the duplicate letters. please Heelllllp
Hi Chris, great question. This is a tricky one that had me thinking for a bit. Consider a) ONOWAY at first. Pretend that the "O"s are different: there’s O1 and O2.
To find regular 4-letter combinations, do
C(6,4) = 15. That assumes the "O"s are different. Because they are the same, we need to subtract duplicate items like
"O1"WAY and "O2"WAY
How many duplicates do we have? Well, we find the number of ways to have an O and some combination of the remaining letters. We need an O + 3 other letters (chosen from 4):
C(4,3) = 4
Once we subtract off the duplicates we get:
C(6,4) - C(4,3) = 15 - 4 = 11
For b), we would do
C(7,4) - C(5,3) = 35 - 10 = 25
I’ll leave c) up to you :).
I have stumped by this one! Can anyone help, please?
Lisa lost the combination to the safe where the
secret cookie recipe is held. She sent for
Bill Becker, the most prolific safecracker
in the prison system, and offered him a
royal pardon if he succeeded in opening
the safe.
After several attempts at bypassing the
combination, Bill realized that the only
way to open the safe is to try every possible
combination by hand. The special lock
has a four-character code. Two of the
characters must be letters, and the lock is
case sensitive (with AB not the same as
ab). The other two must be digits,
anything from 0 to 9.
What is the maximum number of
combinations that Bill would have to
try before finding the correct code?
Thankz kalid, The site is so very cool.I am glad to visit this.
Thanks Giridhar, glad you found it
1.How many different creations can you create all together using one ice cream flavor and at least one mix-in. (there are 52 flavors and 33 mixins)
- How many different combinations of pizzas can you make using at least one topping including crust options. ( 5 crusts, 17 toppings)
my question is the one above! i need help… asap thanks
Thanx a bunch, loved it!
@katie: This sounds a bit like a homework problem; I think I’ll have to do a follow-up on counting techniques.
@Naushad: Awesome, glad it helped
I have a hw problem that seems to involve both permutation and combination. Can you make any suggestions on how to put it all together? :
100 people / 4 prizes; two of “this”, and two of “that”. How many ways to award the prizes if a person “x” wins one of “that”.
So, I see that there are 99 left in the pool, and that two of the prizes of the 3 remaining are the same, so combination is in needed and permutation. But how?
Sorry, THANKS!
Hi Dennis, I’ll take a quick stab. If I understand right, there’s 100 people and two prizes (A and A) and two other prizes (B and B). I assume there’s 1 prize per person.
First, just think about giving out 4 random prizes (A B C and D). You’d just pick 4 winners from 100: P(100,4)
This is a permutation because the order matters – prize A is different from prize D.
However, this doesn’t take the duplicate prizes into account. There’s 2 ways to arrange the Bs. There’s 2 ways to arrange the As. So, we need to divide by 4 to handle these combinations (2 is simple enough no formula is needed, but technically 2 = C(2,1)… how many ways can you pick 1 item (the item to swap) from 2?).
(Note: if all prizes were the same, we’d divide by 4 * 3 * 2 * 1, instead of 4, and end up turning the permutation formula into a combination).
Hope this helps. (And hopefully I didn’t mess it up).
@Katie: I realize I should give you a hint to get started.
For the ice cream, you’re going to get quite a large number. First, you have 52 choices for ice cream.
Next, for each mix-in you can decide to leave it in our out. That is two options per mixin, for 2^33 options total. You need to subtract 1 because you can’t leave all the mixins out. So you’d have something like
52 * (2^33 - 1) which is a pretty large number.
The pizza question would be similar.
Thanks for your time Kalid (on Sunday nonetheless). You know, now with just 3 weeks to go, I can safely say that Discrete Math has presented me with more headache than Linear & DifEq combined…lol.
OK, I’m still uncertain. 1 of four is accounted for. Thus, I understand that if B,C,D were distinct then it would be as simple as P(99,3). From above, my little mind extracted P(99,3)/2 since two of the prizes are the same. Not quite a straight Permutation or Combination??? AHHHH!
P.S. To what ends does this site address. I just found it yesterday, and I’m quite impressed. I enjoyed the explanation above and the view of previous post. I have many friends coming up behind me that I will inform of this site. And as for myself, just 1 left Probability, Stats, & Modeling.
Thanks;
Dennis
Wait a NY minute… if person “x” wins 1 of 4 prizes, and because of duplicates P(4,1)/2 = 2. Do I then get 2*P(99,3)/2 = P(99,3); again dividing by 2 for duplicates?
Thanks.
Hi Dennis, thanks for the comments. Yep, this site is about any topic that has given me or others grief, though usually on math/programming/business/communication topics (as I’m most interested in those).
I think I just thought of an easier way. Suppose we pick the “winners” first and then hand out the prizes. There are C(99,3) ways to pick 3 winners from 99.
Let’s call them 1, 2 and 3. We can distribute the remaining prizes (2As and 1 B) like so:
123
AAB
ABA
BAA
So, we have 3 * C(99,3) possibilities. I think :).
I don’t know what it is, but this subject is not staying in my head. I just don’t get it. I honestly can’t see the difference between the two…I’m going crazy, but I need to learn this stuff. Help!
Hi Shakara, you might have to read this explanation (and others) a couple of times. To me, I think about whether the order I pick people makes a difference. For some things (picking 1st, 2nd, 3rd) the order matters, for other things (just making a group of 3 people) the order doesn’t matter.
If the order matters, then there’s “more ways to pick” since you could have done it one of several ways.
I am having a difficult time with this and I have a test tomorrow. I’ve read several examples but my problems confuse me. My HW asks:
How many ways can a teacher pick four students from a class of 20 to clean up after a party?
How would I do that problem?
Also, how do I compute P(6,3) and C(3,3)?
I’m so lost right now.