abhisheksharma85 wrote:

How many 4 digit numbers can be formed with the digits 0, 1, 2, 3, 4, 5, 6 and 6?

a. 220

b. 249

c. 432

d. 216

e. 288

Dear abhisheksharma85,

I'm happy to comment on this.

What is the source of this question?? From what I can tell, not only is the OA not correct, but the answer I calculate isn't listed among the answer choices and isn't even close.

This is a fascinating counting question. For more on counting, including the

Fundamental Counting Principle, see:

http://magoosh.com/gmat/2012/gmat-quant-how-to-count/Here's how I would approach it. First, I'll ignore the repeat, and count the four-digit numbers I can make with {0, 1, 2, 3, 4, 5, 6}. I am assuming that, in order to be a true four-digit number, zero cannot be in the first digit, the thousands place --- i.e.

0246 does not count as a "four-digit" number.

For the four-digit numbers from {0, 1, 2, 3, 4, 5, 6} ---

For the thousands digit, {1, 2, 3, 4, 5, 6} ---

six choices

For the hundreds digit, drop the one picked in the first choice, but add 0 as a possibility --- still

six choices

For the tens digit, now also drop what was picked in the hundred digit --- now,

five choices

For the ones digit, now also drop what was picked in the tens digit --- now,

four choices

Total number = 6*6*5*4 =

720For example, first choice, from six options {1, 2, 3, 4, 5, 6}, I could choose 3

Then, from these six options {0, 1, 2, 4, 5, 6}, I could choose 0

Then, from these five options {1, 2, 4, 5, 6}, I could choose 2

Then, from these four options {1, 4, 5, 6}, I could choose 1

This produces the unique four-digit number 3021

Now, we have to consider the four-digit numbers with two 6's. First, let's think of where the two 6's could fall among the two digit --- there are 4C2 = 6 possible locations for the two 6's:

(a) 66 _ _

(b) 6 _ 6 _

(c) 6 _ _ 6

(d) _ 66 _

(e) _ 6 _ 6

(f) _ _ 66

I grouped them this way, because in (a)-(c), the thousands digit is already occupied, so zero would be one legitimate choice for either of the other slots, BUT in (d) - (f), we have to be careful, again, not to place zero in the blank in the thousands place.

For each of (a) - (c), we have six choices {0, 1, 2, 3, 4, 5} for the left slot, and then, dropping that one digit, five choices remaining for the right slot. 5*6 = 30 for each of the three, so this results in

90 more numbers.

For each of (d) - (f) we have five choices {1, 2, 3, 4, 5} for the left slot, the thousands place; then, for the right slot, we drop the digit we chose already, but we add zero as a possible choice, and thus we still have five choices. 5*5 = 25 for each of the three, so this results in

75 more numbers.

Altogether 720 + 90 + 75 =

885That's my count of the number of possible four-digit numbers we could form from this set.

Does all this make sense?

Mike

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)