<p>How many positive four_digit integers have 1as their first digit and 2 or 5 as their last digit ?
A)144
B)180
C)200
D)300
E)720
And why????</p>
<p>The answer is (C) 200.
There are 1,000 four digit numbers with their first digit being 1 (1,000 - 1,999)
Out of those, only 2/10 numbers will have 2 or 5 as their last digit (ex. 1002 and 1005)
2/10 * 1000 = 200</p>
<p>I hope this helps :)</p>
<p>Thank you:)</p>
<p>You can also do it like this:</p>
<hr>
<p>Set up four spaces and use permutations.</p>
<p>For space 1 you can only have 1 number (1), so you put a 1.</p>
<p>1 _ _ _</p>
<p>For the middle to you can have 0-10, so you can have 10 numbers for each of the middle slots.</p>
<p>1 10 10 _</p>
<p>And the last slot you can have only 2 or 5, which limits you to 2 numbers for the last one.</p>
<p>1 10 10 2</p>
<p>Now all you need to do is multiply across and you have your answer.</p>
<p>This would not be considered a permutation. But it is a straightforward application of the counting principle.</p>