Combinatorics question (1 Viewer)

[calipress123]

Each of the letters A, A, C, H, M, and S is written on a separate card. The cards are drawn at random from a hat and placed next to each other to form a word. What is the probability that the word ASCHAM appears?

My approach was:
Numerator is 2 as the 2 A's can take the first place and will be correct, and denominator is 6! as it doesn't specify having 'unique' solutions. So 2/6!=1/360. Is this correct?

 

[fromthethickofit]

close but incorrect, give it another shot

 

[calipress123]

The answer says 1/(6!/2!) to give the same answer of 1/360, but I don't get how you do it like that because aren't there 2 possible ways to spell ASCHAM with the A's swapped around still spelling ASCHAM?

 

[funnytomato]

The answer says 1/(6!/2!) to give the same answer of 1/360, but I don't get how you do it like that because aren't there 2 possible ways to spell ASCHAM with the A's swapped around still spelling ASCHAM?

It depends on if the A's are regarded the same or not. By the way it is phrased the 2 A's are identical and hence the 'swapped' version is exactly the same and hence only counted once.

In your way of reasoning if we regard the A's as 'A1' and 'A2' then there are indeed 2 ways with the A's swapped.
Correspondingly the total is 6! as the A's are not identical if using 'A1' and 'A2'

In almost all similar examples just think of a smaller number of objects and literally you can verify it by writing down all arrangements.
E.g. if you had the letters A,A and B what is the probability to get 'ABA'

 

[calipress123]

That makes sense, but I feel like in this situation when drawing the letters randomly from a hat even though the A's are regarded the same they both work as the 1st letter? So like for the 1st letter there's 2 possible letters that they have that can go there - A and A. Or like select a particular A and there's 2 spots for that A so 1 spot for the other A. idk though

 

[ivanradoszyce]

Ask yourself the question.

How many choices (from 6 letters) are there to select the letter A?

Your answer should be .

Ask yourself the question.

How many choices (from 5 letters remaining) are there to select the letter S?

Your answer should be .

Ask yourself the question.

How many choices (from 4 letters remaining) are there to select the letter C?

Your answer should be .

I think you can continue the process to get .

 

[C2H6O]

Each of the letters A, A, C, H, M, and S is written on a separate card. The cards are drawn at random from a hat and placed next to each other to form a word. What is the probability that the word ASCHAM appears?

My approach was:
Numerator is 2 as the 2 A's can take the first place and will be correct, and denominator is 6! as it doesn't specify having 'unique' solutions. So 2/6!=1/360. Is this correct?

This IS correct, because you are speaking as if the a's are unique and so have accounted for that by having 2 ways to spell it

The answer says 1/(6!/2!) to give the same answer of 1/360, but I don't get how you do it like that because aren't there 2 possible ways to spell ASCHAM with the A's swapped around still spelling ASCHAM?

This method treats the A's as the same, so there is only 1 way to spell the word, but the total number of combinations has accounted for the double up by dividing by 2!

 

[calipress123]

This IS correct, because you are speaking as if the a's are unique and so have accounted for that by having 2 ways to spell it


This method treats the A's as the same, so there is only 1 way to spell the word, but the total number of combinations has accounted for the double up by dividing by 2!

Ohhh does that mean they both work??