Question

how many distinct 4-digits integers one can make from the digits 1,3,3,7,7,8 ? Is there and method or we have to go by enumerating the possibilites ?

Answer 1

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!

ADD all these = 4!/2!+4!/2!+4!/2!+4!/2!+4!/2!+4!/2!+4!/(2!*2!)+4! 

                      =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

Answer 2

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

Comments

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!)

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Need more correct approach