retagged by
4,632 views

2 Answers

Best answer
2 votes
2 votes

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

selected by
0 votes
0 votes
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

Related questions

1 votes
1 votes
2 answers
1
Anu asked Apr 28, 2015
3,435 views
How many seven digits number are there such thatDigits are distinct integers taken from {1, 2, ..., 9} andDigits 5 and 6 do not appear together (consecutively)
0 votes
0 votes
1 answer
2
Sankaranarayanan P.N asked Jun 4, 2015
544 views
how many distinct question paper can be set with 10 questions which have four choices and there is only one correct answer per questions.
2 votes
2 votes
1 answer
4