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24 Aug 2019, 04:29
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4 digit no ; aabb
4!/2!*2! ; 6
and the total digits can be ; 2 digit to choose from 9 digits ; 9c2 ; 36
so total 4 digit + integes ; 36* 6 ; 216
IMO D
4!/2!*2! ; 6
and the total digits can be ; 2 digit to choose from 9 digits ; 9c2 ; 36
so total 4 digit + integes ; 36* 6 ; 216
IMO D
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Joined: 05 Jul 2020
Last visit: 26 Dec 2020
Posts: 15
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Given Kudos: 41
Schools: UCLA MSBA "23
14 Sep 2020, 22:25
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The 4-digit number comes in 3 ways: AABB, ABBA, ABAB
For AABB:
Number of options for the first "A" is 9, for the second "A" is 1 since the second "A" has to be the same as the first "A"
Number of options for the first "B" is 8, since it can't be the same as "A." One option for the second "B" since it has to be the same as the first "B"
Therefore, number of options for selecting a "AABB" style 4-digit number is 9*8=72
The same goes for ABBA and ABAB.
Therefore, number of options we have for the 4-digit number is 72*3=216.
For AABB:
Number of options for the first "A" is 9, for the second "A" is 1 since the second "A" has to be the same as the first "A"
Number of options for the first "B" is 8, since it can't be the same as "A." One option for the second "B" since it has to be the same as the first "B"
Therefore, number of options for selecting a "AABB" style 4-digit number is 9*8=72
The same goes for ABBA and ABAB.
Therefore, number of options we have for the 4-digit number is 72*3=216.
16 Sep 2020, 22:30
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to answer the OP as shortly as possible:
you are "Double Arranging" when you set it up that way. I followed your same path, but then realized when you are arranging the 9 different options in the 1 slot and then 8 different options in the other slot, you have already arranged these when you performed the 4!/2!*2! (presumably to find how many ways we can arrange: aabb ---Type of Number)
1st) Number of ways we can arrange the Number of Type: aabb --- in which 2 Digits are the same and a Different 2 Digits are the Same
4! / (2!*2!) = 6 Scenarios
AND
2nd) for Any 1 Arrangement, the Different Combinations of Numbers are the Following:
Assuming Scenario: aa bb
9 Options ----- Locked 1 Option ------ 8 Options Remaining ------ Locked 1 Option
9 * 1 * 8 *1 / 2! ----- Divide out by 2! to avoid over-counting because we do NOT want to count the Arrangements in which the a and b switch. We already did that in Step 1.
Answer:
(4! / 2!*2!) * (9 * 1 * 8 * 1 / 2!) =
6 * (72/2) =
6 * 36 = 216 Numbers that follow the constraints.
The Other Method is to 1st Choose which 2 Distinct Digits are going to be the Repeats that show up in the 4 Digit Number
1st) Selection/Choosing
You can Choose which 2 Distinct Digits that will Repeat in the Number in the following ways:
"9 choose 2" Ways = 36 ways
AND
2nd)ARRANGEMENT
For Each Combination of 2 Distinct Digits Chosen, We can Arrange them in the following ways:
4! / 2! * 2! ------ because 2 of the Elements will be Indistinguishable Elements (2a 2b 3 3) is the same Arrangment as (2b 2a 3 3)
36 * 6 = 216 Ways
-D-
you are "Double Arranging" when you set it up that way. I followed your same path, but then realized when you are arranging the 9 different options in the 1 slot and then 8 different options in the other slot, you have already arranged these when you performed the 4!/2!*2! (presumably to find how many ways we can arrange: aabb ---Type of Number)
1st) Number of ways we can arrange the Number of Type: aabb --- in which 2 Digits are the same and a Different 2 Digits are the Same
4! / (2!*2!) = 6 Scenarios
AND
2nd) for Any 1 Arrangement, the Different Combinations of Numbers are the Following:
Assuming Scenario: aa bb
9 Options ----- Locked 1 Option ------ 8 Options Remaining ------ Locked 1 Option
9 * 1 * 8 *1 / 2! ----- Divide out by 2! to avoid over-counting because we do NOT want to count the Arrangements in which the a and b switch. We already did that in Step 1.
Answer:
(4! / 2!*2!) * (9 * 1 * 8 * 1 / 2!) =
6 * (72/2) =
6 * 36 = 216 Numbers that follow the constraints.
The Other Method is to 1st Choose which 2 Distinct Digits are going to be the Repeats that show up in the 4 Digit Number
1st) Selection/Choosing
You can Choose which 2 Distinct Digits that will Repeat in the Number in the following ways:
"9 choose 2" Ways = 36 ways
AND
2nd)ARRANGEMENT
For Each Combination of 2 Distinct Digits Chosen, We can Arrange them in the following ways:
4! / 2! * 2! ------ because 2 of the Elements will be Indistinguishable Elements (2a 2b 3 3) is the same Arrangment as (2b 2a 3 3)
36 * 6 = 216 Ways
-D-
