<p>1- How many positive 3-digit integers can be formed using only the digits 3 and 4 if each of these digits must be used at least once in each 3-digit integer?</p>
<p>A) 4
B) 6
C) 8
D) 10
E) 12</p>
<p>And please don't count, we all know how to count :P But what I'm looking for here, is the model answer/approach to this question. For some reason, my math tutor says that the general counting principle of "how many possibilities" does not apply here. Why?</p>
<p>Simplest way is just:
2 choices for the first digit (you can choose 3 or 4)
2 choices for the second digit (you can choose 3 or 4)
2 choices for the third digit (you can choose 3 or 4)
2^3 = 8 combinations</p>
<p>But you can’t have 333 or 444, which means you have two less combinations than the original formula.</p>
<p>8 - 2 = 6</p>
<p>At least I hope thats it.</p>
<p>Well, your tutor is right that it is probably easier to list them…but here’s what I think you are looking for:</p>
<p>For each of the three digits, you have two choices. That’s 2 x 2 x 2 =8.</p>
<p>But then you have to subtract the two numbers that don’t have one of each digit: 333 and 444. So there are 8 - 2 = 6.</p>
<p>But I would have just listed and counted…</p>
<p>8-2 = 6 is great.</p>
<p>I would’ve counted to check my answer anyway. (I’m too insecure :()</p>
<p>You’re all right. </p>
<p>Emphasis on the “both 3 and 4 must be used in the number”. Forgot to subtract. Thanks once again to CC!</p>
<p>Okay, I understand what you guys did, but WHY doesn’t the counting principle apply here? I have tried multiplying 2<em>2</em>1 for two possibilities (3 and 4) times another two possibilities ( 3 and 4 ) then times one possibility ( either 3 or 4 ). But of course that gives 4, not 6.</p>
<p>We DID use the counting principle. That’s where 2x2x2 came from. Then we subtract the two cases that didn’t use both numbers. And that’s what you were kind of doing as well. You even said “either 3 or 4”. That’s TWO possibilities, not one. So it looks like you got tangled up trying to account for the fact that you had to use one of each. It can be a good strategy in counting problems to count more than you wanted and then figure out what to subtract. It’s kind of like the shaded-region of counting :)</p>