The number of all 3-digit numbers abc for which abc + ab+bc + ac+a+b+c = 29 The number of all 3-digit numbers abc (in base 10) for which $\ abc+ab+bc+ac+a+b+c = 29$ is
(A) 6
(B) 10
(C) 14
(D) 18  
My working:
$\ ab (c + 1) +b (c + 1) + a(c + 1) + c + 1 = 30$
$\ (a + 1) (b +1) (c+1) = 30 $
$\ 9>a>0$
$\ 0\le b,c<9$   
The problem I am facing:
I don't know how to figure out the no. of solutions for which this is true.

$$(a+1)(b+1)(c+1)=30$$
Total number of ways to satisfy above equation:


*

*$30 = (2,3,5)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=1, b=2, c=4)$

*$30 = ( 1 , 6 ,5)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=0, b=5, c=4)$

*$30 = (1,3,10)$ gives $6$
    values for $a+1,b+1,c+1$ $(i.e., a=0, b=2, c=9)$



In the first case we have $6$ values for $a,b,c$ as $(1,2,4)$ can take up any digit hence they can be arranged in $3!$ ways.
In the second case we have $4$ values for $a,b,c$ as $(0,5,4)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2.2.1$ number of ways .
Similarly in the third case we have $4$ values  $a,b,c$ as $(0,2,9)$.  $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number (  contrary to given condition ). Hence we have $ 2.2.1$ number  of ways .
Therefore, total number of ways:
$$3!+2\times 2\times1+2\times 2\times1=6+4+4=14 \space \text{ways}$$
