How many random 3 letter words have exactly 1 “A”? No repetitions. Trying to build intuition here
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I think the options are either:
c(3,1) * c(25,2) (as in, choose 1 of the three letters to be the “A” and choose 2 of the remaining 25 letters for the other one) = 900
OR
if the first letter is an A, then there are 25 options for the second and 24 for the third. There are 3 slots for the A, so 25*24*3 = 1800
Are either correct? If so, what’s the intuition around why 1 is wrong?
combinatorics permutations combinations
asked Dec 7 '18 at 19:39
ARBARB
82
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add a comment |
$begingroup$
I think the options are either:
c(3,1) * c(25,2) (as in, choose 1 of the three letters to be the “A” and choose 2 of the remaining 25 letters for the other one) = 900
OR
if the first letter is an A, then there are 25 options for the second and 24 for the third. There are 3 slots for the A, so 25*24*3 = 1800
Are either correct? If so, what’s the intuition around why 1 is wrong?
combinatorics permutations combinations
asked Dec 7 '18 at 19:39
ARBARB
82
$endgroup$
add a comment |
$begingroup$
I think the options are either:
c(3,1) * c(25,2) (as in, choose 1 of the three letters to be the “A” and choose 2 of the remaining 25 letters for the other one) = 900
OR
if the first letter is an A, then there are 25 options for the second and 24 for the third. There are 3 slots for the A, so 25*24*3 = 1800
Are either correct? If so, what’s the intuition around why 1 is wrong?
combinatorics permutations combinations
asked Dec 7 '18 at 19:39
ARBARB
82
$endgroup$
I think the options are either:
c(3,1) * c(25,2) (as in, choose 1 of the three letters to be the “A” and choose 2 of the remaining 25 letters for the other one) = 900
OR
if the first letter is an A, then there are 25 options for the second and 24 for the third. There are 3 slots for the A, so 25*24*3 = 1800
Are either correct? If so, what’s the intuition around why 1 is wrong?
combinatorics permutations combinations
combinatorics permutations combinations
asked Dec 7 '18 at 19:39
ARBARB
82
asked Dec 7 '18 at 19:39
ARBARB
82
asked Dec 7 '18 at 19:39
ARBARB
82
asked Dec 7 '18 at 19:39
ARBARB
82
asked Dec 7 '18 at 19:39
ARBARB
82
82
add a comment |
add a comment |
1 Answer
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You forgot to permute the two remaining letters in the first approach. They will permute in $2!$ ways, so the total ways will be multiplied by $2!$.
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
$endgroup$
1
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You forgot to permute the two remaining letters in the first approach. They will permute in $2!$ ways, so the total ways will be multiplied by $2!$.
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
$endgroup$
1
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
add a comment |
$begingroup$
You forgot to permute the two remaining letters in the first approach. They will permute in $2!$ ways, so the total ways will be multiplied by $2!$.
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
$endgroup$
1
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
add a comment |
$begingroup$
You forgot to permute the two remaining letters in the first approach. They will permute in $2!$ ways, so the total ways will be multiplied by $2!$.
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
$endgroup$
You forgot to permute the two remaining letters in the first approach. They will permute in $2!$ ways, so the total ways will be multiplied by $2!$.
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
answered Dec 7 '18 at 19:43
Shubham JohriShubham Johri
5,057717
5,057717
1
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
add a comment |
1
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
1
1
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
$begingroup$
ahhh. makes sense. thanks!
$endgroup$
– ARB
Dec 7 '18 at 19:52
add a comment |
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