<p>So, recently on my previous SAT practice test, I bumped into a fill-in permutation question. It roughly stated that "In a four digit number, the first number must be two and the last number cannot be ten. If numbers may be repeated, how many different digits can you think of." I don't remember what the exact question was, taking into consideration that I took this test three days ago.</p>
<p>Do you mean zero instead of 10 as a last digit? 'Cause…</p>
<p>2XY10 is not four digits.</p>
<p>If so, there are 900 possibilities.</p>
<p>The last number cannot be 10? The digits are 0 to 9.</p>
<p>I do not understand your question since one digit cannot be ten anyway…
Do you mean, that the last two digit cannot be 10?</p>
<p>If so:
Just multiply the possible events for each digit.
The first one has to be 2. So there is only one possibility.
The second digit can be, whatever. So there are 10 possibilities ranging
from 0 to 9. If the last two digits cannot be ten in the number then there are two possibilities: The third digit is not 1 or the last digit is not 0.</p>
<p>So 1 x 10 x 9 (no 1) x 10 + 1 x 10 x 1 x 9 (no 0)
IF this is correct, there should be 990 different numbers.</p>
<p>Can somebody else confirm this or correct me? It seems quite logical to me.
You could also approach it vice-versa. </p>
<p>If the first digit must be a two, there are exactly 1000 possible numbers which comply with this requirement. Since the last two (?) numbers cannot be ten, digits one, three, and four are fixed: 2x10 (whereas x can range from 0 to 9)
There are only 10 numbers out of the 1000 possible numbers, which does not fit the requirements. Therefore, 990 should be correct.</p>
<p>Sorry. Yes, I mean to say ZERO not ten.</p>
<p>Okay, its even easier then.
First digit = 2 => one possibility
Second digit = 0 to 9 => ten possibilities
Third digit = 0 to 9 => ten possibilities
Fourth digit = 1 to 9 (no 0) => nine possibilities</p>
<p>Multiply all the possibilities.
1 x 10 x 10 x 9 = 900</p>
<p>There are 900 possible numbers that fulfill the requirements.
It’s easy, isn’t it?</p>
<p>P.S. Nevertheless, can somebody confirm my approach for the question if the last two numbers are not 10; is it correct?</p>
<p>Did you take the international SAT or the USA one? I somehow vaguely remember this question, but I was sure it didn’t stipulate a nonzero number for the ones digit. Anyways, your approach is correct because you can repeat numbers.</p>
<p>Thanks!
By the way, he said that this question appeared on a practice test and not on a real one! ;)</p>
<p>^ I agree with your answer.</p>
<p>You could also do it this way:</p>
<p>If you don’t care about the last 2 digits, there are 1x10x10x10 = 1000 ways. But some of them do end in 10. How many? 1 x 10 x 1x 1 = 10. We don’t want those…So 1000-10 = 990.</p>