There is a 3 digit positive number such that sum of all its digits is equal to 6. How many such number are possible?

Question

A four digit positive number having all non-zero distinct digits is such that the product of all the digits is least. If the difference of hundreds digit and tens digit is 1.
How many possibilities are there of such number?

Answer 1

sum of digits == 6

,105,114 ,123 , 132 , 141, 150 ,

204,213 ,222 , 231 ,,240,

303, 312 ,,321, 330 ,

402 , 411 , 420,

501 , 510 ,

600

Hence total numbers are ====21

Answer 2

1,0,5,these three number can be arrange by 3!=6,since 015 and 051 is not valid hence =4
1,1,4,can be arranged by 3!/2=3
1,2,3, can be arranged by 3!=6
2,2,2 can be arranged by 3!/3!=1
2,0,4 can be arranged by 3!-2=4
3,0,3 can be arranged by 3!/2!-1=2
6,0,0 can be arranged by 1!=1
4+3+6+1+4+2+1=21 is the answer