1. ## Combination Question about words with repeated letters

1)In how many ways is it possible to select one or more letters from those in INSIPIDITY?

2)In how many ways is it possible to select&nbsp; six letters including at least one vowel FLABELLIFORM?

FOR 1) I have tried to use the formula for combinations with repetitions $\displaystyle frac{(n+r-1)!}{r!(n-1)!}$ but I have not able to get the answer 319
I thought; since it is one or more letters then I should consider each case; such as if i select one letter then two letter then 3 letters etc then I did 10C1 + 10C2; but the I dont know how to account for the repeated 4 I's
for 2)since it has at least one vowel we have two methods; the first method total - none ( not selecting vowels)
normally it is see to do this12C6 - 8C6; but it is not right due to the complexity of the repeated letters f, f, l, l, l
so; i use the second method count the selection of choosing 6 letters with one vowel, two vowels, three vowels; and four vowels

one vowel (4C1 = 4);
vowels: a e i o& ;consonants: f f l l l b r m
i) one f;
f (a/e/i/o) _ _ _ _6C4 for selecting non-vowels.
4C1 x 6C4 = 60
ii) 1 f 1 L
f (a/e/i/o) L _ _ _
6C3 for selecting non-vowels. 4C1 x 6C3 = 4 x 20 = 80
iii) 2 f 3 L
f (a/e/i/o) f L L L
only choosing one vowel 4C1 = 4

iv)2 f 1 L;f (a/e/i/o) f L _ _
6C4 for selecting non-vowels.
4C1 x 6C2 = 60
v) 2 f 2 L;<br>&nbsp;f (a/e/i/o) f L L _
6C4 for selecting non-vowels. 4C1 x 6C1 = 24

two vowel (4C2 = 6);
i) one f;
f (a/e/i/o) (a/e/i/o)_ _ _
4C2 x 6C3 = 120
if i was to add it but to now;
60 + 60+80+24+4+120 = 348 but the answer is 296
The problem i have is that i believe that I am on the right track meaning that i believe i have to consider each different condition such as choosing 1 f 0 l 1 vowels and other consonants;
Please I would like some help with this thank you

2. ## Re: Combination Question about words with repeated letters

1. EDITED: I realize this doesn't come out to 319. I wonder what I'm missing ... can you post the entire problem statement?

Keep a running "list" of possible outcomes and analyze your possibilities based on what's in this list.
How many ways are there to select 1 letter from INSIPIDITY? There are 7 unique letters: I, N, S, P, D, T, Y and if you pick one at a time, then there are 7 possible selections. Put each possible outcome in your list: (I, N, S, P, D, T, Y)

How many ways are there to select 2 letters from INSIPIDITY? Again -- 7 unique letters.
Suppose you first pick I, how many possible selections are left? Since you can't re-pick I (because you don't count duplicates and don't replace the letter you've chosen), there are 6 selections left: (IN, IS, IP, ID, IT, IY). So if you can only select 2 unique letters at a time and your first pick was I, then there are 6 possibilities left. Put those 6 possibilities in your list: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY). What if you had picked N first? Then there are 6 selections possible: (NI, NS, NP, ND, NT, NY) -- add them to your list BUT notice that NI and IN are 'the same outcome' (since ordering doesn't matter). The list you now have should be: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY, NS, NP, ND, NT, NY). What if you had picked S first? Then, if we again list the possible outcomes AND remove the permutations we already have in the list you get: (SP, SD, ST, SY). Add those 4 outcomes to your list because you don't have them yet. What if you had picked P first? D? T? Y? (PD, PT, PY), (DT, DY), (TY).

Your list of outcomes should be: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY, NS, NP, ND, NT, NY, SP, SD, ST, SY, PD, PT, PY, DT, DY, TY). Notice that 7 of those outcomes came from constraining yourself to select 1 unique letter and the other 21 came from constraining yourself to select 2 unique letters. Hopefully, you can see that by constraining ourselves to pick only 'k' unique letters from 7 possible unique letters, there is exactly "7 choose k" possible outcomes: 7 choose 1 = 7, 7 choose 2 = 21. Our list of possible outcomes for k=1 and k=2 is 7 + 21 = 28 possible outcomes.

How many unique ways are there to select 3 letters? 4? 5? 6? 7? 8? 9? (of course, beyond 7, you ran out of unique letters so n choose k, for k>n is always 0). How many outcomes is in your list after counting all of the unique ways? I count 127.

3. ## Re: Combination Question about words with repeated letters

Originally Posted by Elusive1324
1. EDITED: I realize this doesn't come out to 319. I wonder what I'm missing ... can you post the entire problem statement?

Keep a running "list" of possible outcomes and analyze your possibilities based on what's in this list.
How many ways are there to select 1 letter from INSIPIDITY? There are 7 unique letters: I, N, S, P, D, T, Y and if you pick one at a time, then there are 7 possible selections. Put each possible outcome in your list: (I, N, S, P, D, T, Y)

How many ways are there to select 2 letters from INSIPIDITY? Again -- 7 unique letters.
Suppose you first pick I, how many possible selections are left? Since you can't re-pick I (because you don't count duplicates and don't replace the letter you've chosen), there are 6 selections left: (IN, IS, IP, ID, IT, IY). So if you can only select 2 unique letters at a time and your first pick was I, then there are 6 possibilities left. Put those 6 possibilities in your list: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY). What if you had picked N first? Then there are 6 selections possible: (NI, NS, NP, ND, NT, NY) -- add them to your list BUT notice that NI and IN are 'the same outcome' (since ordering doesn't matter). The list you now have should be: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY, NS, NP, ND, NT, NY). What if you had picked S first? Then, if we again list the possible outcomes AND remove the permutations we already have in the list you get: (SP, SD, ST, SY). Add those 4 outcomes to your list because you don't have them yet. What if you had picked P first? D? T? Y? (PD, PT, PY), (DT, DY), (TY).

Your list of outcomes should be: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY, NS, NP, ND, NT, NY, SP, SD, ST, SY, PD, PT, PY, DT, DY, TY). Notice that 7 of those outcomes came from constraining yourself to select 1 unique letter and the other 21 came from constraining yourself to select 2 unique letters. Hopefully, you can see that by constraining ourselves to pick only 'k' unique letters from 7 possible unique letters, there is exactly "7 choose k" possible outcomes: 7 choose 1 = 7, 7 choose 2 = 21. Our list of possible outcomes for k=1 and k=2 is 7 + 21 = 28 possible outcomes.

How many unique ways are there to select 3 letters? 4? 5? 6? 7? 8? 9? (of course, beyond 7, you ran out of unique letters so n choose k, for k>n is always 0). How many outcomes is in your list after counting all of the unique ways? I count 127.
I have tried that but it is still do not add up to 319

4. ## Re: Combination Question about words with repeated letters

Not that I don't trust you but it's very important that you state the problem exactly as written. If you reinterpreted the question and accidentally omitted some important detail, then "319" might have answered a different question!
I'm going to assume you did state the problem as written. Let's consider a simpler example, BEEF, since it's easier to list the possibilities to check if our logic holds. It contains duplicates and has 3 unique letters. The unique letters are (B, E, F)

We ask: How many possible ways are there to select one or more letters?

Total number of ways to select one or more letters = total number of ways to select one unique letter + total number ways to select two unique letters + total number of ways to select three unique + ...

Let's list the outcomes, underlining all duplicate outcomes:

total number of ways to select one unique letter: (B, E, F)
total number ways to select two unique letters: (BE, BF, EB, EF, FB, FE)
total number of ways to select three unique letters: (BEF, ...)

That's 3 ways to select one letter, 3 ways to select 2 letters, 1 way to select 3 letters (all other permutations are equivalent to 'BEF'). Thus the total is 7.
Of course, this is (3choose1) + (3choose2) + (3choose3) = 7.

Working with this smaller example tells me that my method is reasonable since I can directly enumerate the possible outcomes. The smaller example has features of the bigger word: it contains a single duplicated letter.

The only way it *isn't* 7 is if some of those duplicated outcomes counted, which means there was a caveat to the question that would have included them ... which means the question being asked was different.

5. ## Re: Combination Question about words with repeated letters

Originally Posted by bigmansouf
1)... I dont know how to account for the repeated 4 I's
If the problem statement tells you to count repeated letters, then you can label them i, i_1, i_2, i_3. (underscore indicates subscript). If it doesn't then there is only 1 letter 'i'.

6. ## Re: Combination Question about words with repeated letters

Originally Posted by bigmansouf
1)In how many ways is it possible to select one or more letters from those in INSIPIDITY??
Both of these are extremely poorly written questions with respect to counting theory.
In counting the word select usually means content alone not in any way does it imply order.

Using the set $\{I,N,S,P,D,T,Y\}$ there are $2^7-1=127$ nonempty subset of that set. Therefore, that is the number of selections of at least one.
Now, remove the $I$. There are $2^6-1=63$ selections which contain at least one of $\{N,S,P,D,T,Y\}$
Suppose we add the empty set to that new collection. That gives us $64$ unique subsets.
There are $\dbinom{64-1+4}{4}=\dbinom{67}{4}=766480$ ways to place four identical $I's$ into $64$ unique sets.

@bigmansouf, please repost the exact wording of these questions. What is the source of these questions?

7. ## Re: Combination Question about words with repeated letters

Originally Posted by Plato
Both of these are extremely poorly written questions with respect to counting theory.
In counting the word select usually means content alone not in any way does it imply order.

Using the set $\{I,N,S,P,D,T,Y\}$ there are $2^7-1=127$ nonempty subset of that set. Therefore, that is the number of selections of at least one.
Now, remove the $I$. There are $2^6-1=63$ selections which contain at least one of $\{N,S,P,D,T,Y\}$
Suppose we add the empty set to that new collection. That gives us $64$ unique subsets.
There are $\dbinom{64-1+4}{4}=\dbinom{67}{4}=766480$ ways to place four identical $I's$ into $64$ unique sets.

@bigmansouf, please repost the exact wording of these questions. What is the source of these questions?
the first question is number 21
the second question is number 26

thank you

also please can you explain how you got $2^7 -1 = 127$ from
What I am getting to understand is why you choose to raise 2 by 7 and not choose 3 or 4 or 5.

thank you

8. ## Re: Combination Question about words with repeated letters

Originally Posted by bigmansouf
the first question is number 21
the second question is number 26

also please can you explain how you got $2^7 -1 = 127$ from
What I am getting to understand is why you choose to raise 2 by 7 and not choose 3 or 4 or 5.
Your images are side-ways for me and almost impossible to read. (it my age, I know that)!
Let me tell you that I know that you live in the UK. But I have taught there while my wife lectured on a Fulbright.
Moreover, I wrote and edited questions in counting theory for years. I have never before encountered the wording of this question.
Here is my example, using the wording of your question.
How many ways is it possible to select one or more letters from those in $\{S,O,U,T,H,E,R,N\}$
That is a set of eight unique letters. A basic count of the number of subsets of a set of $N$ unique elements is $2^N$.
Now one of those subsets is the empty set, $\emptyset$. So there are $2^N-1$ non-empty subsets.
Thus there are $2^8-1$ non-empty subsets of $\{S,O,U,T,H,E,R,N\}$ that is also the number of possible to selections of one or more letters from those in $\{S,O,U,T,H,E,R,N\}$.
NOW consider the string AOOUROOMBOTOON. That is a string of fourteen letters half of which are unique the other seven are O's.

So consider the question "How many ways is it possible to select one or more letters from" that string?
Well there are eight elements in $\{A,O,U,R,M,B,T,N\}$ so there $2^8-1=255$ non-empty subsets.
But that does not include any set the has more than one $O$.
So lets do this: there are $2^7=128$ subsets of $\{A,U,R,M,B,T,N\}$ each of which is unique (including the empty set).
Now we have seven identical $Os$ to distribute into $128$ unique cells(subsets).
The number of ways to put $N$ identical objects into $k$ unique cells is $\dbinom{N+k-1}{N}$
Put seven identical $Os$ into $128$ unique subsets of $\{A,U,R,M,B,T,N\}$ in $\dbinom{128+7-1}{7}=5169379425$

Now I must say again. I have no idea what this question means to ask.

9. ## Re: Combination Question about words with repeated letters

INSIPIDITY

has 4 I's and 6 other letters.

Suppose we wanted to find out the number of distinct, say 5 letter combos

There are 5 possibilities, 0-4 I's.

see if you can follow the next line

$n_k = \dbinom{6}{5-k},~k=0,1,\dots 4$

we pick $k$ Is. There's only one way to do it. Then we pick the remaining letters from the 6.

Now we want the total number of combos of any length, 1 to 10

$N = \sum \limits_{n=1}^{10}\sum \limits_{k=0}^{\min(n,4)}\dbinom{6}{n-k} = 319$

really not that big a fuss.

10. ## Re: Combination Question about words with repeated letters

FLABELLIFORM

(FF)(LLL)BRM(AIEO)

and you want 6 letter combos that include at least 1 vowel.

This is just a tedious exercise in counting.

you'll have
0, 1, or 2 Fs
0, 1, 2, or 3 Ls

select these. There will only be one way to do this.

select your vowel, there are $\dbinom{4}{1}$ ways to do this

you then have 6 letters remaining to choose from. Choose your remaining letters from these.

I'm gonna let you do the hard work on this one.

11. ## Re: Combination Question about words with repeated letters

Your solution suggests that the problem counted combinations containing different numbers of I’s as distinct. Is that right? For example: if n=3, NII and III are possible outcomes that are counted by the inner summation when k=2 and k=3 respectively.

You gotta admit the question was pretty vague regarding how to treat duplicate i’s. Without knowing that the author wanted 319 as the solution, one could have interpreted the question differently and arrive at 127.