Problem Solving

How many three-digit numbers have at least one 2 and at least one 3?

(A) 52
(B) 54
(C) 56
(D) 58
(E) 60

I do not have the OA. Please post your answers with approach.

I got the answer as A) 52

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.

I am just learning permutations, if i have committed any mistakes kindly excuse and spot and inform any mistakes if so. Provided there is no OA the explanation needs validation.

I got D) 58, but would be interested in the official answer. here is my method:

at least two digits need to be a 3 or a 2, which I will use the place holder O for, so there are three potential formations (with the X being the spot that can be any digit including 2 or 3).

1) X,O,O
2) O,X,O
3) O,O,X

Now for the Hundreds digit we can not have a zero there, so there are 9 possible numbers that could go there, times the fact that the 3 and the 2 could be in either the ten or the single digit spot:

1) 9,O,O times 2 = 18

For the tens and single digit there are ten possible numbers that can go there, times the fact that the 3 and the 2 can again go in either spot:

2) O,10,O times 2 = 20
3) O,O,10 times 2 = 20

Adding these possibilities up gives us: 18+20+20 = 58

Not positive on this, so let me know what others think

sportcntr3
let me further elaborate your approach
1) X,O,O
2) O,X,O
3) O,O,X

1)X, O , O
can be
X 2 3
X 3 2
( i have omitted X 2 2 and X 3 3 , since they wont be repeated, let us take the above two case alone)

2) O, X, O
can be
2 X 3
3 X 2

3) O,O,X
can be
2 3 X
3 2 X

when counting by your approach the number 223 from 1)X 2 3 is counted again in 2) 2 X 3
Similarly the number 323 from 1) X 2 3 is counted again in 3) 3 2 X
Like wise six numbers gets counted twice in this approach. If you further look carefully can spot other numbers that are counted again also. Hope am right..

crejoc you are correct, I forgot the initial permutation you need to run to eliminate the overlap, so 3!/2! (to find the overlapping instances of two 3's) + 3!/2! (to find the overlapping instances of two 2's) = 6

So then 58 - 6 (duplicate numbers for the times X is a 2 or a 3) = 52

The numbers ranges from 123 to 923
x23: 9 numbers (x range from 1 to 9)
x32: 9 numbers (x range from 1 to 9)
23x: 10 numbers (x range from 0 to 9)
32x: 10 numbers (x range from 0 to 9)
2x3: 10 numbers (x range from 0 to 9)
3x2: 10 numbers (x range from 0 to 9)
--------------
Total: 58 numbers => answer: D

vpmba2009 wrote:The numbers ranges from 123 to 923
x23: 9 numbers (x range from 1 to 9)
x32: 9 numbers (x range from 1 to 9)
23x: 10 numbers (x range from 0 to 9)
32x: 10 numbers (x range from 0 to 9)
2x3: 10 numbers (x range from 0 to 9)
3x2: 10 numbers (x range from 0 to 9)
--------------
Total: 58 numbers => answer: D

vpmba2009
by your approach 6 numbers are counted twice leading to the answer of 58, read the previous posts for clarity the answer is 52.

sportcntr3 wrote:crejoc you are correct, I forgot the initial permutation you need to run to eliminate the overlap, so 3!/2! (to find the overlapping instances of two 3's) + 3!/2! (to find the overlapping instances of two 2's) = 6

So then 58 - 6 (duplicate numbers for the times X is a 2 or a 3) = 52

yes , it is always better to deal with this kind of problems by splitting the problem for each case.