Q. Find the total no of 10 digits whose sum is 4.

• its 220..
  4=4+0(one 4 & 9 zeroes)=> 1
  4=3+1(one 3,one 1 & 8 zeroes)=> 2*9!/8!=18
  4=2+2(two 2's & 8 zeroes)=> 9
  4=1+1+2(two 1's & one 2)=> 9!/7! + 9!/(7!*2!)=72+36=108
  4=1+1+1+1(four 1's)=> 9!/(6!*3!)=84
  total no. of 10 digits no. whose sum is 4 =1+18+9+108+84=220
  ans 220
• 10 years agoHelpfull: Yes(29) No(4)
• 4=3+1=====>2*9!/8! i cant understand this combination plx explain
• 10 years agoHelpfull: Yes(9) No(0)
• 1) 4=4+0==>there is only one chance==>1
  2) 4=3+1==>if first number is 3,then remaining numbers can be arranged in 9! of which 8 are zeros i.e.,repeated numbers==>9!/8!.....similarly if the first number is 1,i.e.,9!/8!===>2*9!/8!=18
  3) 4=2+2===>if 2 is first number then remaining can be arranged as 9!/8!==>9
  4) 4=1+1+2===>if first number is 1 then remaining 9 numbers can be arranged in 9! of which 7 are
  zeros==>9!/7!....and if first number is 2 then remaining 9 numbers can be arranged as 9! of which 7 zeros,2 ones.===>9!/(7!*2!)===>9!/7!+9!/(7!*2!)===72+36=108.
  5) 4=1+1+1+1===>if first number is 1 then remaining 9 numbers can be arranged as 9! of which 6 zeros,3 ones.==>9!/(6!*3!)=84.
  so,total no. of 10 digits no. whose sum 4=1+18++9+108+84=220(answer)
  
• 9 years agoHelpfull: Yes(5) No(0)
• 4=4+0(one 4 & 9 zeroes)=> 1
  4=3+1(one 3,one 1 & 8 zeroes)=> 2*9!/8!=18
  4=2+2(two 2's & 8 zeroes)=> 9
  4=1+1+2(two 1's & one 2)=> 9!/7! + 9!/(7!*2!)=72+36=106
  4=1+1+1+1(four 1's)=> 9!/(6!*3!)=84
  total no. of 10 digits no. whose sum is 4 =1+18+9+106+84=218
  ans 218
• 10 years agoHelpfull: Yes(4) No(16)
• consider 10 digits as a+b+... as given sum=4.a+b+c+d....=4.as given in questions sum of digits .so digits may be0,1,...9.now as, a+b+c+...=4 so no. of +ve solutions=n+s-1cn-1=13c9=715
• 10 years agoHelpfull: Yes(2) No(1)
• Naeem can u plz explain all this...
  4=3+1(one 3,one 1 & 8 zeroes)=> 2*9!/8!=18
  4=2+2(two 2's & 8 zeroes)=> 9
  4=1+1+2(two 1's & one 2)=> 9!/7! + 9!/(7!*2!)=72+36=108
  4=1+1+1+1(four 1's)=> 9!/(6!*3!)=84...
  how this combination is done?
  
• 10 years agoHelpfull: Yes(2) No(0)
• 4=1+1+1+1 or 2+2 or 1+1+2,
  Four 1's can be arranged in 10 places in 10C4 ways and two 1's and one 2 can be arranged in 10C3, also 1,1,2 can be arranged in 3 different ways and two 2's can be arranged in 10C2 ways, So total number of 10 digits whose sum is 4= 10C4+10C3*3+10C2 which is equal to 615.
• 10 years agoHelpfull: Yes(1) No(5)
• A- spanish, french
  B-spanish,english,french
  C-spanish,english,italian,porchigese
  D-spanish,
  E-spanish,french,italian,english,porchigese
• 8 years agoHelpfull: Yes(0) No(1)