Latest Infosys Aptitude Question SOLUTION: Q. Find the total no of 10 digits whose sum is 4.

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

Asked In Infosys BISWAJIT (10 years ago)

Solved Naeem Ahamad Read Solution (8)

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Naeem Ahamad  (10 years ago)

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

Is this solution Helpfull? Yes (29) | No (4)  

praveensiva  (10 years ago)

4=3+1=====>2*9!/8! i cant understand this combination plx explain

Is this solution Helpfull? Yes (9) | No   

bhanu raj  (9 years ago)

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)

Is this solution Helpfull? Yes (5) | No   

RAKESH  (10 years ago)

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

Is this solution Helpfull? Yes (4) | No (16)  

rohitashwa  (10 years ago)

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

Is this solution Helpfull? Yes (2) | No (1)  

Smruti Sudha Mishra  (10 years ago)

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?

Is this solution Helpfull? Yes (2) | No   

Piyush Kumar  (10 years ago)

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.

Is this solution Helpfull? Yes (1) | No (5)  

P.R.PRATHAP  (8 years ago)

A- spanish, french
B-spanish,english,french
C-spanish,english,italian,porchigese
D-spanish,
E-spanish,french,italian,english,porchigese

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