@IIMAIM said:Find the Number of ways of forming a 4 digit number using 1,2,3 ?
3C1 +3C2(4!/3! +4!/2!2!) +3C3(4!/2!)
3+3(4+6)+12
45?
all digits same 3c1*4!/4!=3
3 digits same 1 different 3c1*2c1*4!/2!2!=24
2 digits same 2 different 3c1*2c1*4!/2!2!=18
adding all 3+24+18=45
@iLoveTorres said:whats the approach then?
1st 2 digits are repeated
1 1 2 3
1 1 3 2
2 2 1 3
2 2 3 1
3 3 1 2
3 3 2 1
similarly 2nd and 3rd digit are repeated,3-4,1-3,1-4,2-4
Total=6*6=36
1 1 2 3
1 1 3 2
2 2 1 3
2 2 3 1
3 3 1 2
3 3 2 1
similarly 2nd and 3rd digit are repeated,3-4,1-3,1-4,2-4
Total=6*6=36
@IIMAIM said:1st 2 digits are repeated 1 1 2 3 1 1 3 2 2 2 1 32 2 3 13 3 1 23 3 2 1similarly 2nd and 3rd digit are repeated,3-4,1-3,1-4,2-4Total=6*6=36
sar ke upar se gaya@abhishek.2011 said:all digits same 3c1*4!/4!=33 digits same 1 different 3c1*2c1*4!/2!2!=242 digits same 2 different 3c1*2c1*4!/2!2!=18adding all 3+24+18=45
we have to use all 1,2,3 in the number so just one number repeats
@IIMAIM said:check my approach
bhai where are the cases when all digits are same
2 same 2 different
3 same 1 different
also 2 same 2 same
@IIMAIM said:1st 2 digits are repeated
1 1 2 3
1 1 3 2
2 2 1 3
2 2 3 1
3 3 1 2
3 3 2 1
similarly 2nd and 3rd digit are repeated,3-4,1-3,1-4,2-4
Total=6*6=36
did u include the case like 1122 2233 3311 ????
@Subhashdec2 said:bhai where are the cases when all digits are same 2 same 2 different 3 same 1 differentalso 2 same 2 same
bhai my question is we have to use 1,2,3 in the number and just 1 number has to repeat
@IIMAIM said:1st 2 digits are repeated
1 1 2 3
1 1 3 2
2 2 1 3
2 2 3 1
3 3 1 2
3 3 2 1
similarly 2nd and 3rd digit are repeated,3-4,1-3,1-4,2-4
Total=6*6=36
All the cases are not considered. Look for all the digits same and 3 digits same.
@abhishek.2011 said:did u include the case like 1122 2233 3311 ????
my question says u have to use all 1,2,3 in the four digit number so just one number repeats
cases when all are same 1111 2222 3333 = 3
cases when 3 are same 1112/3 2221/3 3331/2 3*4!/3!=12
cases when 2 are same and next 2 are different 1123 2213 3312 = 3*4!/2!=18
cases when 2 are same and next 2 are also same 1122 2233 3311 = 3*4!/2!*2!=18
18+18+12+3=51
@vijay_chandola said:All the cases are not considered. Look for all the digits same and 3 digits same.
question says u have to consider all 1,2,3 in four digit number generated
@IIMAIM
ur question says number of 4 digit numbers that can be formed,its nowhere written that u have to use all the digits??
@IIMAIM said:question says u have to consider all 1,2,3 in four digit number generated then how could u take 3 digits same?
if the case is you can have 1,2, or all the digits same isnt this equivalent to each number can be treated in 4 ways so the three numbers can be treated in 3^4 ways?
@IIMAIM said:question says u have to consider all 1,2,3 in four digit number generated then how could u take 3 digits same?
ok now do this question in this way
how many 4 digit numbers can be formed with 1,2,3 (any digit can repeat itself any number of times)
@abhishek.2011 said:@IIMAIM ur question says number of 4 digit numbers that can be formed,its nowhere written that u have to use all the digits??
I have edited the question.