@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?
@abhishek.2011 said:ok now do this question in this wayhow many 4 digit numbers can be formed with 1,2,3 (any digit can repeat itself any number of times)
@iLoveTorres said:why have you considered the number of digits as 9? 0-9=10 right? now subtracting the case of 0 you get 9 possibilities
@vijay_chandola said:Question is like this: 'Find the Number of ways of forming a 4 digit number using 1,2,3 ?'Nowhere it is stated that we have to use 'all' the digits.
@scrabbler said:Oh that's not number of digits. That is the sum of digits. I have a sum of digits of 9, which has to be distributed into the 4 available "boxes" or "groups" such that the first box is not empty...regardsscrabbler
@IIMAIM said:Find the Number of ways of forming a 4 digit number using 1,2,3 in every number generated ?
@Logrhythm said:when all digits are same -> 3c1 = 3 numbers...when three digits are same -> 3c1*4!/3! = 12 numbers..when any 2 digits are same and other 2 digits are different -> 3c1*4!/2! = 36 numbers...when 2 digits are same and other 2 digits are also same -> 3c1*2c1*4!/2!^2 = 36 numbers...total = 3+12+36*2 = 87?? -----------------------ok it says 1,2,3 in every number generated means we need to use 1,2 and 3 in every number..when 2 digits are same and other 2 are different -> 36 numbers...so is the answer 36 or 87?? @IIMAIM
@Logrhythm said:when all digits are same -> 3c1 = 3 numbers...when three digits are same -> 3c1*4!/3! = 12 numbers..when any 2 digits are same and other 2 digits are different -> 3c1*4!/2! = 36 numbers...when 2 digits are same and other 2 digits are also same -> 3c1*2c1*4!/2!^2 = 36 numbers...total = 3+12+36*2 = 87??-----------------------ok it says 1,2,3 in every number generated means we need to use 1,2 and 3 in every number..when 2 digits are same and other 2 are different -> 36 numbers...so is the answer 36 or 87?? @IIMAIM
@Logrhythm said:when all digits are same -> 3c1 = 3 numbers...when three digits are same -> 3c1*4!/3! = 12 numbers..
@Logrhythm said:when any 2 digits are same and other 2 digits are different -> 3c1*4!/2! = 36 numbers...when 2 digits are same and other 2 digits are also same -> 3c1*2c1*4!/2!^2 = 36 numbers...
@Logrhythm said:total = 3+12+36*2 = 87??
@Logrhythm said:-----------------------ok it says 1,2,3 in every number generated means we need to use 1,2 and 3 in every number..when 2 digits are same and other 2 are different -> 36 numbers...
@Logrhythm said:so is the answer 36 or 87?? @IIMAIM
@scrabbler said:This should be 3C1 * 2C1 * 4!/3! = 24 since we also have to choose which of the other 2 digits is chosen to appear once.This again should be divided by 2 as 2 same and 2 same means I am just choosing which 2 out of the 3 make an appearance which is 3C2 ways and not 3C1 x 2C1 so 18 total.Hence 3 + 24 + 36 + 18 = 81.Or of course we could do the 3^4 case.This is only the 3rd case above. Hence 36.In fact, if you want to nitpick, it still can be interpreted differently (after the correction) as to whether other digits besides 1, 2, 3 are allowed or not. So there could also be another set of cases whether 1,2,3 and some other digit are used. These would be 7 * 4! - 3! (using any of the other 7 available digits, and then removing the numbers which would start with a 0) in addition to the 36 above = 168 - 6 + 36 = 198 numbers which have a 1, 2 and 3 present.Question setters need to take loads of care to ensure there is no ambiguity They rarely do...regardsscrabbler
Now that we are discussing PnC....i have had a doubt everytime i attempt such question. I am sure u guys can help...
@Logrhythm said:Now that we are discussing PnC....i have had a doubt everytime i attempt such question. I am sure u guys can help...How can 7 balls be distributed among 3 earns when:1) balls are similar and earns are not2) earns are similar but balls are not3) both balls and earns are similar i end up getting the right answer but after 1 or 2 hit and trials as i am always doubtful of my approach.......PS - If you any more variants of such questions.....feel free to educate..
@IIMAIM said:Thanks scrabbler , yes there has been some ambiguity ,I am apologizing for that.But an important question to discuss, in which u have explored all the possible ways of asking it.
@Logrhythm said:Now that we are discussing PnC....i have had a doubt everytime i attempt such question. I am sure u guys can help...How can 7 balls be distributed among 3 earns when:1) balls are similar and earns are not(i) when empty earn is allowed (ii) empty earn not allowed2) earns are similar but balls are not(i) when empty earn is allowed (ii) empty earn not allowed3) both balls and earns are similar (i) when empty earn is allowed (ii) empty earn not allowedi end up getting the right answer but after 1 or 2 hit and trials as i am always doubtful of my approach.......PS - If you any more variants of such questions.....feel free to educate..
@Logrhythm said:Now that we are discussing PnC....i have had a doubt everytime i attempt such question. I am sure u guys can help...How can 7 balls be distributed among 3 earns when:1) balls are similar and earns are not(i) when empty earn is allowed (ii) empty earn not allowed2) earns are similar but balls are not(i) when empty earn is allowed (ii) empty earn not allowed3) both balls and earns are similar (i) when empty earn is allowed (ii) empty earn not allowedi end up getting the right answer but after 1 or 2 hit and trials as i am always doubtful of my approach.......PS - If you any more variants of such questions.....feel free to educate..
@maroof10 said:1.If a,b,c,d are in continued proportion then a-d/b-c>=x. What is the value of x?2.If a,b,c,d are proportional then the mean proportion between a^2+c^2 and b^2+d^2 is ?PLZ HELP