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