crejoc's Explanation:
There are three possibilities for a three-digit numbers to have at least one 2 and at least one 3
They are
1) having two '2' and one '3' eg. 223
2) having two '3' and one '2' eg. 332
3) having one '2' , one '3' and one 'any other number than 2 and 3' eg. 235
consider each case separately,
In the first case,
1) there are three possible numbers 223, 232, 322
we can also find by the usual permutation method to find arrangements of letters of a word
eg.) different arrangements for word 'BOB' is 3!/2! = 3 , 3! for 3 letters and 2! for similar letters.
In the second case,
2) this condition can also be worked out as option 1, so 3 numbers
In the third case,
3) the third condition is different, the possible numbers are
23x
2x3
32x
3x2
x32
x23
x being any other number excluding 2 and 3.
now let us consider 1 by 1
23x==>1*1*8 = 8, 1st position can be filled by no.2 only, 2nd position can be filled by no.3 only, third position can be filled by 8 numbers excluding 2 and 3 from possible 10 numbers ,
2x3 ==>8( as above)
32x ==>8
3x2 ==>8
x32 ==>7
here it is 7 because 1 st position cannot be a zero, so excluding 2,3 and 0 from 10 possible numbers.
x23 ==>7(as above)
so for the third case 8+8+8+8+7+7=46
sum up all the three cases 3+3+46= 52 is the answer.
Source :
https://www.beatthegmat.com/how-many-3-d ... 37909.html
