Combinations with identical and unidentical objects

[tellmesomething]

Homework Statement

Find number of ways of selection of any number of letters from AADEF if there are no restrictions

Relevant Equations

!!

MY ATTEMPT: Case 1: No. of ways of selecting 0 letters-1
Case 2: No of ways of selecting 1 letter: From identical A's- 1
From D E F- 3C1= 3
Total=4
Case 3: No. of ways of selecting 2 letters: From identical A's- 1
From D E F - 3C2= 3
1 A and D or E or F= 3
Total-7
Case 4: No. of ways of selecting 3 letters: 2A's and 1 D or E or F- 3
1 A and DE or EF or FD- 3
From D E F-1
Total-7
Case 5: No. of ways of selecting 4 letters: 2As and DE or EF or FD :3
1A and DEF: 1
Total: 4


Case 6: NO. of ways of selecting 5 letters - AADEF: 1
Total=24 I know my logic is flawed somewhere that is why the answer is not matching but where i cant pinpoint please help me out.PS: I know theres a neater way to do this with a formula but i do not understand why it works so im trying to atleast convince myself it works by brute forcing it like this. So please spot the mistake here and do not use the formula to arrive at the ans :)

Merry christmas!!

 

[PeroK]

First, in terms of brute force you can simply write them all down and then count them. Second, 24 looks right to me.

One way to do this is first to consider the four letters AADE without the F. Now, if we include the F, then all those selections are valid, plus each one again with the F. So:
$$N(AADEF) = 2 \times N(AADE)$$And, by the same logic:$$N(AADEF) = 8 \times N(AA)$$Finally, $N(AA) = 3$ if we consider the A's identical and $N(A_1A_2) = 4$ if we consider the A's different.

 

[PeroK]

PS this also gives you a method for listing all the possibilities. I'll use "-" for the selection:
-
A
AA
D
AD
AAD

E
AE
AAE
DE
ADE
AADE

Then also those 12 with an F.

 

[WWGD]

Homework Statement: Find number of ways of selection of any number of letters from AADEF if there are no restrictions
Relevant Equations: !!

MY ATTEMPT: Case 1: No. of ways of selecting 0 letters-1
Case 2: No of ways of selecting 1 letter: From identical A's- 1
From D E F- 3C1= 3
Total=4
Case 3: No. of ways of selecting 2 letters: From identical A's- 1
From D E F - 3C2= 3
1 A and D or E or F= 3
Total-7
Case 4: No. of ways of selecting 3 letters: 2A's and 1 D or E or F- 3
1 A and DE or EF or FD- 3
From D E F-1
Total-7
Case 5: No. of ways of selecting 4 letters: 2As and DE or EF or FD :3
1A and DEF: 1
Total: 4


Case 6: NO. of ways of selecting 5 letters - AADEF: 1
Total=24 I know my logic is flawed somewhere that is why the answer is not matching but where i cant pinpoint please help me out.PS: I know theres a neater way to do this with a formula but i do not understand why it works so im trying to atleast convince myself it works by brute forcing it like this. So please spot the mistake here and do not use the formula to arrive at the ans :)

Merry christmas!!

Please look up multinational coefficients. This addresses cases such as the number of permutations of " Mississippi ".

 

[haruspex]

multinational

multinomial?

 

[haruspex]

I also confirm 24, thus:

• distinguishing all five, $2^5=32$
• that counts each combination with exactly one A twice
• so half of the 32 are counted twice
• 32-16/2=24

 

[WWGD]

multinomial?

Yes, that too;) .

 

[WWGD]

Well, the Multinomial coefficient, designed for this situation gives us $\frac {5!}{2!}=60$, corresponding to setting one $A$ as $A_1$, another $A:= A_2$, and collapsing symmetric arrangements such as $A_1A_2DEF, A_2A_1DEF$ . Thing is, with 4 different letters we can form $4!=24$ combinations. Wouldn't an additional $A$ add something to the total?
Though it depends on how/when we define two arrangements to be equal.

 

[haruspex]

Thing is, with 4 different letters we can form 4!=24 combinations.

No, $2^4$.

 

[WWGD]

No, $2^4$.

Well, it would depend on how you identify equal strings, whether we allow replacement, etc. I fail to see how the total isn't 4!: 4 choices for 1st, three for the second, etc.

 

[haruspex]

Well, it would depend on how you identify equal strings, whether we allow replacement, etc. I fail to see how the total isn't 4!: 4 choices for 1st, three for the second, etc.

That is wrong in two ways. First, that procedure means you have to select four. Secondly, you have counted each combination (of which there is only one in that procedure) 24 times.
The way to think of it is that each of the four is either selected or not.

 

[WWGD]

That is wrong in two ways. First, that procedure means you have to select four. Secondly, you have counted each combination (of which there is only one in that procedure) 24 times.
The way to think of it is that each of the four is either selected or not.

Yes, I was referring to how many ways there are of choosing 4 , from the set $\{A,E,D,F\}$. I'm aware if you choose any amount, you get $\Sigma C_{n,i}; i=1,2,..,n$, which equals $(1+1)^n=2^n$.

 

[haruspex]

Yes, I was referring to how many ways there are of choosing 4 , from the set $\{A,E,D,F\}$.

To which the answer is 1.

 

[WWGD]

To which the answer is 1.

Again, this depends on the notion, criterion, you use to decide when two strings are equivalent. _If_ ordering doesn't matter, then, yes, AEDF=AFED=..... But _If_ order does matter, then there are $24$ such strings.EDIT If you're assigning rooms to 4 people, $\{A,E,D,F\}$, where $(A,E,D,F)$ is the assignment of A to room 1, B to room 2, etc. , then there are $24$ such assignments.

 

[haruspex]

Again, this depends on the notion, criterion, you use to decide when two strings are equivalent. _If_ ordering doesn't matter, then, yes, AEDF=AFED=..... But _If_ order does matter, then there are $24$ such strings.

The question asks for the number of selections of any number of ... It is quite standard that that implies the order is irrelevant.

 

[WWGD]

@tellmesomething : Seems you've been gone for a while. Can you clarify here the conditions of the problem?

 

[haruspex]

@tellmesomething : Seems you've been gone for a while. Can you clarify here the conditions of the problem?

From post #1, it seems @tellmesomething interpreted the question the same way I do. OTOH, it implies 24 is not the official answer.
The only alternative I see is to take order as significant. In that case I get 162 by sheer counting.
For n distinct objects, the number of order-significant sequences of 0 or more is $\lfloor e\cdot n!\rfloor$. Haven't figured out how to modify that when two are the same.