+1 vote
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if all digits are distinct so {1,3,7,8}=4! ways

only two digits are same so :

five  cases :

case 1 : {1,3,3,8} =4!/2!

case 2:{1,7,7,8}=4!/2!

case 3:{3,3,7,7}=4!/(2!*2!)

case 4:{1,3,3,7}=4!/2!

case 5:{8,1,7,7}=4!/2!

case 6:{8,3,3,7}=4!/2!

case 7;{1,7,7,3}=4!/2!

=102 ans

this above method is sometimes not so good but with this way we can also do this question ! :)

but the better approach is :

You have to try all cases:

Case 1: all distinct digits: 4!=24

Case 2: two same digits and two different digits:

that implies two more sub-cases:

(i) {3,3,-,-} these two blanks can only be filled by two out of the three {7,1,8} = C(3,2)=3 and then arrangement of these 4 numbers = (4!/2!); In total by pdt rule=3*12=36

(ii) {7,7,-,-} similarly here 2 out of the three {3,1,8}=3 and then arrangement=(4!/2!). total by pdt rule=3*12=36;

Case 3: two same and other two same:

{3,3,7,7}=(4!/(2!*2!))=6

by sum rule, total possibilities=6+2*36+24=102

selected
Number of distinct elements=1,3,7,8

1st digit:4 ways

2nd digit:4 ways

3rd digit:4 ways

4th digit :4 ways

Tot: 4*4*4*4 ways
+2
I have got the answer. Sorry, but you are not trying all possible cases.

You have to try all cases:

Case 1: all distinct digits: 4!=24

Case 2: two same digits and two different digits:

that implies two more sub-cases:

(i) {3,3,-,-} these two blanks can only be filled by two out of the three {7,1,8} = C(3,2)=3 and then arrangement of these 4 numbers = (4!/2!); In total by pdt rule=3*12=36

(ii) {7,7,-,-} similarly here 2 out of the three {3,1,8}=3 and then arrangement=(4!/2!). total by pdt rule=3*12=36;

Case 3: two same and other two same:

{3,3,7,7}=(4!/(2!*2!))=6

by sum rule, total possibilities=6+2*36+24=102
0
I don't think I have missed  cases.  Try my formula taking small size example
0
can you tell me your logic ? what made you conclude that all have 4 ways and therefore, pow(4,4);

suppose you have {1,3,3,7} and you have to make 2-digit distinct numbers ? Are you saying with the above approach, that unique={1,3,7} and therefore both places has 3 and 3 possibilities, which makes total=9 ?
0
and if I try to break down this problem into cases:

case 1: all distinct means I have {1,3,7}, and to make 2-digit distinct, I can possibly do permutation P(3,2)=6.

Case 2: two same digits :{3,3} , only one possibility =1

Total= 6+1=7;

Also, If you enumerate it like set with repeated elements={1,3,3,7}; and you ought to make 2-element ordered set then, they are : {1,3} ,{3,1},{1,7},{7,1},{3,7},{7,3},{3,3}=7.
0
Sorry I was wrong! You have to enumerate it With 1,3,7,8 4! Possibility With 3,3,7,7 4!/2!2! Possibility With 1,3,3,8. 4!/2!2! Possibility With 1,7,7,8 4!/2!2! Possibility With 1,3,3,7 4!/2! Possibility With 1,7,7,3. 4!/2! Possibility Anything missing ??
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I think possibilities with (1,3,3,8 ) would be 4!/(2!) and same is for (1,7,7,8). you are missing these :

(3,3,7,8) = 4!/(2!) and (3,7,7,8) = 4!/(2!)
0
Need more correct approach

+1 vote