Digits

[jfranco23]

Of the 3 digit positive integers that have no digits equal to 0, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?

A. 24
B. 36
C. 72
D. 144
E. 216

[codesnooker]

IMO E

Lets say XYZ is a three digit number.

Since zero is not allowed at any of the place,

So, for X we have 9 choices, for Y we have 8 choices and for Z we have only 1 choice.

Total options = 9 * 8 * 1 = 72

Now since, each number can placed at three different position, so total choices will be 72 * 3 = 216

[jfranco23]

how you get that y is 8 and z is 1?

[cubicle_bound_misfit]

Of the 3 digit positive integers that have no digits equal to 0, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?

A. 24
B. 36
C. 72
D. 144
E. 216




we have three places say ----- ------ ------
A B C

how many ways can you fill place A --- 9 ways
now either B or C is equal to A

if B is eqal total number of permutation : 9*1*8
If C is equal total no of permutation 9*8*1

only another case remains that is if BC equal A different

so total, 3(9*8*1) == 216.

Waht is OA?

[crackgmat007]

[quote="jfranco23"]Of the 3 digit positive integers that have no digits equal to 0, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?

A. 24
B. 36
C. 72
D. 144
E. 216[/quote][color=darkblue][/color]

IMO, using Box approach is simpler for computing permutations. In the question, we have 3 digits. Lets say 3 boxes. Since the question states that 2 digits are equal, for computation purpose lets consider the 2 equal digits as one box. This gives us 2 boxes. In the first box, one of the 9 numbers can be assigned. This leaves the next box that can be assigned with 8 numbers.

Hence 9*8=72

The 2 equal digits can be placed in two ways e.g 989, 899 (note that the third way of placing is fliping the first combination, 989, is redundant). Multiply the above with 2; hence 72*2=144.

Answer is D - IMO.