How many strings of four decimal digits do not contain the same digit twice
0
$begingroup$
The answer should be
1: Four different digits : $$10*9*8*7 = 5040 $$
2: Four different digits + contains the same digit three times + all digits are of the same : $$ 10*9*8*7+10*{4 choose 3} *9 + 10 = 5410 $$
Which one is correct? Depend on different interpretation?
discrete-mathematics permutations combinations
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
$endgroup$
add a comment |
0
$begingroup$
The answer should be
1: Four different digits : $$10*9*8*7 = 5040 $$
2: Four different digits + contains the same digit three times + all digits are of the same : $$ 10*9*8*7+10*{4 choose 3} *9 + 10 = 5410 $$
Which one is correct? Depend on different interpretation?
discrete-mathematics permutations combinations
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
$endgroup$
1
$begingroup$
I think yes, it opens up to interpretation. Depending on what you interpreted, both answers are correct. I presume 0123 is also considered as a 4 decimal digit number
$endgroup$
– Icycarus
Dec 7 '18 at 11:20
1
$begingroup$
I would interpret the problem the first way since the strings $1332$, $3331$, and $3333$ contain the digit $3$ (at least) twice.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 11:21
1
$begingroup$
If a digit occurs thrice, it occurs twice too.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:25
1
$begingroup$
If the second way is what was intended, the question should be phrased as "... exactly twice" in my opinion. Therefore I think the first way is correct.
$endgroup$
– Erik André
Dec 7 '18 at 11:28
add a comment |
0
0
0
$begingroup$
The answer should be
1: Four different digits : $$10*9*8*7 = 5040 $$
2: Four different digits + contains the same digit three times + all digits are of the same : $$ 10*9*8*7+10*{4 choose 3} *9 + 10 = 5410 $$
Which one is correct? Depend on different interpretation?
discrete-mathematics permutations combinations
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
$endgroup$
The answer should be
1: Four different digits : $$10*9*8*7 = 5040 $$
2: Four different digits + contains the same digit three times + all digits are of the same : $$ 10*9*8*7+10*{4 choose 3} *9 + 10 = 5410 $$
Which one is correct? Depend on different interpretation?
discrete-mathematics permutations combinations
discrete-mathematics permutations combinations
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
asked Dec 7 '18 at 11:16
TimgascdTimgascd
303
303
1
$begingroup$
I think yes, it opens up to interpretation. Depending on what you interpreted, both answers are correct. I presume 0123 is also considered as a 4 decimal digit number
$endgroup$
– Icycarus
Dec 7 '18 at 11:20
1
$begingroup$
I would interpret the problem the first way since the strings $1332$, $3331$, and $3333$ contain the digit $3$ (at least) twice.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 11:21
1
$begingroup$
If a digit occurs thrice, it occurs twice too.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:25
1
$begingroup$
If the second way is what was intended, the question should be phrased as "... exactly twice" in my opinion. Therefore I think the first way is correct.
$endgroup$
– Erik André
Dec 7 '18 at 11:28
add a comment |
1
$begingroup$
I think yes, it opens up to interpretation. Depending on what you interpreted, both answers are correct. I presume 0123 is also considered as a 4 decimal digit number
$endgroup$
– Icycarus
Dec 7 '18 at 11:20
1
$begingroup$
I would interpret the problem the first way since the strings $1332$, $3331$, and $3333$ contain the digit $3$ (at least) twice.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 11:21
1
$begingroup$
If a digit occurs thrice, it occurs twice too.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:25
1
$begingroup$
If the second way is what was intended, the question should be phrased as "... exactly twice" in my opinion. Therefore I think the first way is correct.
$endgroup$
– Erik André
Dec 7 '18 at 11:28
1
1
$begingroup$
I think yes, it opens up to interpretation. Depending on what you interpreted, both answers are correct. I presume 0123 is also considered as a 4 decimal digit number
$endgroup$
– Icycarus
Dec 7 '18 at 11:20
$begingroup$
I think yes, it opens up to interpretation. Depending on what you interpreted, both answers are correct. I presume 0123 is also considered as a 4 decimal digit number
$endgroup$
– Icycarus
Dec 7 '18 at 11:20
1
1
$begingroup$
I would interpret the problem the first way since the strings $1332$, $3331$, and $3333$ contain the digit $3$ (at least) twice.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 11:21
$begingroup$
I would interpret the problem the first way since the strings $1332$, $3331$, and $3333$ contain the digit $3$ (at least) twice.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 11:21
1
1
$begingroup$
If a digit occurs thrice, it occurs twice too.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:25
$begingroup$
If a digit occurs thrice, it occurs twice too.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:25
1
1
$begingroup$
If the second way is what was intended, the question should be phrased as "... exactly twice" in my opinion. Therefore I think the first way is correct.
$endgroup$
– Erik André
Dec 7 '18 at 11:28
$begingroup$
If the second way is what was intended, the question should be phrased as "... exactly twice" in my opinion. Therefore I think the first way is correct.
$endgroup$
– Erik André
Dec 7 '18 at 11:28
add a comment |
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1
$begingroup$
I think yes, it opens up to interpretation. Depending on what you interpreted, both answers are correct. I presume 0123 is also considered as a 4 decimal digit number
$endgroup$
– Icycarus
Dec 7 '18 at 11:20
1
$begingroup$
I would interpret the problem the first way since the strings $1332$, $3331$, and $3333$ contain the digit $3$ (at least) twice.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 11:21
1
$begingroup$
If a digit occurs thrice, it occurs twice too.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:25
1
$begingroup$
If the second way is what was intended, the question should be phrased as "... exactly twice" in my opinion. Therefore I think the first way is correct.
$endgroup$
– Erik André
Dec 7 '18 at 11:28