@Brooklyn said:Find the sum of all possible five-digit numbers that can formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.with appraoch
@YouMadFellow
@Brooklyn said:Find the sum of all possible five-digit numbers that can formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.with appraoch
_ _ _ _ 1 (1 occurs here 4*4*3*2 times, similarly others)
When 1_ _ _ _ (It occurs 5*4*3*2 times here)
=> Sum = (1+2+ 3+4 +5)*[ 1 + 10 + 100 + 1000]*[4*4*3*2] PLUS
(1 + 2+ 3 + 4 +5)*[10000] * [5*4*3*2]
= Answer .. (19449856) ? 
Calculation !
Calculation ! @Brooklyn said:Find the sum of all possible five-digit numbers that can formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.with appraoch
The numbers can come in the ten thousands' place 120 times (5!) and in the remaining places 4*4! =96 times. So 10000*15*120 +(1111)*15*96. Kuch 1938000 ya kuch tha ans.
koi yeh waala session attend kar rha hain kya...??
@YouMadFellow said:_ _ _ _ 1 (1 occurs here 4*4*3*2 times, similarly others)When 1_ _ _ _ (It occurs 5*4*3*2 times here)=> Sum = (1+2+ 3+4 +5)*[ 1 + 10 + 100 + 1000]*[4*4*3*2] PLUS(1 + 2+ 3 + 4 +5)*[10000] * [5*4*3*2]= Answer .. (19449856) ? Calculation !
calc to sahi mei wrong hai 😛
ur multiplyin wi 10000 how can last digit be non zero??? :O
PS datgs nt d prob:
dont we do:
(1+2+3+4+5)*4!*11111 - (1+2+3+4+5)*3!*1111 ??
@ankita14 said:The numbers can come in the ten thousands' place 120 times (5!) and in the remaining places 4*4! =96 times. So 10000*15*120 +(1111)*15*96. Kuch 1938000 ya kuch tha ans.
@Brooklyn said:calc to sahi mei wrong haiur multiplyin wi 10000 how can last digit be non zero???PS datgs nt d prob:dont we do:(1+2+3+4+5)*4!*11111 - (1+2+3+4+5)*3!*1111 ??
@bluechameleon18
@Brooklyn said:Find the sum of all possible five-digit numbers that can formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.with appraoch
@Brooklyn said:calc to sahi mei wrong haiur multiplyin wi 10000 how can last digit be non zero???PS datgs nt d prob:dont we do:(1+2+3+4+5)*4!*11111 - (1+2+3+4+5)*3!*1111 ??
Ye Kya hai :/
@ad18 said:Why did you assume that the shadow of the lamppost coincides with the shadow of the boy? Is it the general assumption?
You can ignore dispersion of light at the edges of the "head of the boy" .. The boy is an opaque object, right ? .. :splat:
@Brooklyn said:direct method, u dont knw ???
Kitna method yaad rakhenge jo aaram se ho raha hai uske liye formula kyu yaad rakhu.
jo aaram se ho raha hai uske liye formula kyu yaad rakhu.
@Brooklyn bhai formula hai iska when ever 0 is there in this kinda problems the formula will be (n-2)! [(sum of digits) * {[(n-1)*111...ntimes] - (111...(n-1)times}
@ankita14 said:Kitna method yaad rakhenge jo aaram se ho raha hai uske liye formula kyu yaad rakhu.
mil gyi error :banghead:
@Brooklyn said:calc to sahi mei wrong haiur multiplyin wi 10000 how can last digit be non zero???PS datgs nt d prob:dont we do:(1+2+3+4+5)*4!*11111 - (1+2+3+4+5)*3!*1111 ??
Haan Bhai.. meri calculation thodi weak hai unfortunately.. Answer sahi hai par .. agar thik calculate karo to.. exam me main options ki help leta..
I never thought motivation could come from anything external. But these 2 performances seemed genuine and did move me.
Syvester Styllone: http://www.youtube.com/watch?v=_euvjz0kHLo
@Brooklyn said:Find the sum of all possible five-digit numbers that can formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.with appraoch
Sum = (Sum for all 5 digit no.s using 0,1,2,3,4,5) - (Sum for all 4 digit no.s using 1,2,3,4,5)
(5 + 4 + 3 + 2 + 1 + 0)(11111)(5*4*3*2) - (5+4+3+2+1)(1111)(4*3*2)
@karanthepagal said:@Brooklyn bhai formula hai iska when ever 0 is there in this kinda problems the formula will be (n-2)! [(sum of digits) * {[(n-1)*111...ntimes] - (111...(n-1)times}
i also used this only counted no's as 5 :banghead:
@Angadbir said:I never thought motivation could come from anything external. But these 2 performances seemed genuine and did move me.
Syvester Styllone: www.youtube.com/watch?v=_euvjz...Sum = (Sum for all 5 digit no.s using 0,1,2,3,4,5) - (Sum for all 4 digit no.s using 1,2,3,4)= 19599840(5 + 4 + 3 + 2 + 1 + 0)(11111)(5*4*3*2) - (5+4+3+2+1)(1111)(4*3*2)
yahi kiya tha sir counted digits as 5 :banghead: