<p>Sometimes it's faster to count what you DON'T want...</p>
<p>Example: How many ways are there to form a six-person committee from a group of 7 people?</p>
<p>Yes, it's 7C6. But instead of thinking of it as choosing 6 people you want, think of it as choosing the one person you DON'T want...that makes it obvious that there are 7 ways to do this.</p>
<p>Harder Example: How many 4-letter codes can you make using the first 8 letters of the alphabet if it is required that there must be at least two letters which are used more than once?</p>
<p>And a little harder (to me, anyway): Allen, Bob, Cathy, Dave, Edward and Fran are to be seated in a row of 6 seats. How many ways can we do this if Edward and Fran refuse to sit together?</p>
<p>Bump…and look at the next thread for hints</p>
<p>The third one I think it’s 20:</p>
<p>The pattern is each time we move one of them forward, the possibilities acceptable is minus 1. The first option (Edward first) has 4 possibilities. The second (Edwards sits second) has three, etc. etc.</p>
<p>E - - - - F
E - - - F -
E - - F - -
E - F - -
- E - - - F
- E - - F -
- E - F - -
-
-
-
-
- E - F
4+3+2+1 = 10
And the other way around, so 20</p>
<p>^But each of these that you have counted allows for lots of ways to arrange the others. When you take that into account, your appoach will work.</p>
<p>I was thinking of a different method. Start by thinking of how many total arrangements…Then subtract the ones you DON’T want…</p>
<p>@pckeller - that’s exactly what you would do for #3. You’d rock 6! - all the possibilities.</p>
<h1>3.</h1>
<p>Total possibilities = 6! = 720;
Now let’s find the total number of possibilities in which every other one is the same:
EF----;
-EF—;
–EF–;
—EF-;
----EF;
There are 5;
Multiply by 2 to allow for EF/FE arrangements;
Multiply by 4! or 24 to find the other blank’s possibilities;
The number of combinations we don’t want = (5)(2)(24) = 240;</p>
<p>720 - 240 = 480;</p>
<p>The answer [should be] 480, if my calculations are correct.</p>
<p>Harder ex.
How many 4-letter codes can you make using the first 8 letters of the alphabet if it is required that there must be at least two letters which are used more than once?
8P3 = 336;
T.f. the possible blanks — have 336 combinations;
Now multiply by 3P1 or 3, the repeated letter’s possible values;
336*3 = 1008; This represents the arrangement —n;
I THINK n can be in any of four positions, though, as in: n—, -n–, --n-, or —n.
So the final answer is (336)(3)(4) = 4032.</p>
<p>I’m not confident about either answer, so feedback would be nice.</p>
<p>480 is definitely correct. Working backwards (figuring out which combinations YOU don’t want) is the way to go here. I also remember I’ve seen this particular question somewhere…not sure, might be de ja vu.</p>
<p>I did the same calculations but I was unsure and decided not to post them and humiliate myself if I turn out to be wrong . :D</p>
<p>Hints…</p>
<p>When I write these problems, I try to stay in the spirit of the SAT. So that means you can solve them with the counting principle only and no need for any more advanced combinatorics (such as nPr or nCr).</p>
<p>Start by finding how many 4 letter codes you can make from the first 8 letters, assuming you are allowed to repeat letters as many times as you like.</p>
<p>Then, find how many codes you can make if you are NOT allowed to repeat any letters at all.</p>
<p>If you subtract the second one from the first, you’ll have what we are looking for.</p>
<p>BTW this would have been more SAT-like if I had used smaller numbers, say 3 -letter codes from the first 5 letters…small enough to list and count, big enough so that listing and counting is a bit of a pain.</p>
<p>Guys why so many steps?
5i=120 x2 for FE/EF =240
720-240=480. I know it’s what you were doing but I thought it could be quicker.</p>
<p>1- The first question is the easiest one (obviously) and you solved it in the best way.</p>
<p>2- About the second one, my solution:
First think about the situation, only one: Two letters appear twice respectively.
Calculation:
1) Letter choosing: 8C2=28
2) Array (when the first letter settles its two seats, the rest one is automatically settled, so let the whatever letter out of the two choose firstly): 4C2=6
3) Since the aforementioned two steps are completed step by step, multiply the two results: 8C2 x 4C2 = 168</p>
<p>3- The reasoning process is a little bit difficult if you wanna get the optimal method:
Reasoning:
1) To know how manny arrangements there are that make E and F sit separately, you can use 6P6 minus number of arrangements in which E and F are juxtaposed. (You have used this method in the easiest one.)
2) To calculate how many arrangements E and F sit together, you can first combine them (My high school teacher told us this method of thinking. BTW I come from Mainland China… QUQ)
Calculating:
a) Numbers of arrangements in which E and F sit next to each other:
1) Make E and F a combination (cuz this is about arrangement, EF and FE is different): 2P2=2
2) Since E and F are combined, you can now see there are 5 seats now and you now should place the rest four and the combination in the seats: 5P5=120
3) Multiply the results: 2P2 x 5P5 = 240
b) Find all the possible arrangements: 6P6=720
c) Subtract the arrangements in a) from the possibilities in b): 720-240 = 480</p>
<p>The succinct answers:
1- 7C1 = 7
2- 8C2 x 4C2 = 168
3- 6P6 - 2P2 x 5P5 = 480</p>
<p>Note: I really don’t think question 2 and 3 will appear in SAT I test… Really…</p>
<p>Guys, why are we using permutations on 2? That question is a repitition so permutations don’t work.</p>
<p>It’s 168.
8C3 is 56 (the numbers chosen) then 3C1 (the repeated number) so 56x 3. I could be wrong.</p>
<p>I recalculated the challenge and it’s actually 4032*(3/4) because only 3 of the 4 possible combinations: n—, -n–, --n-, and —n will be unique (because the ints get repeated)!</p>
<p>The answer should be 3024.</p>
<p>e.g. if your first three are 123 and your repeated number is -1-, your combinations will be as follows:
-1-123
1-1-23
12-1-3
123-1-
Notice that the first two possibilities are the same</p>
<p>Either I don’t know how to solve my own problem, or this is the longest time I have ever stumped you guys. Maybe some of you are out of town this weekend :)</p>
<p>If my solution IS correct, you do NOT need permutations or combinations…</p>
<p>Just keep asking, one letter at a time: “How many choices do I have now for THIS letter?” First do that when you ARE allowed to repeat letters (as many times as you want) and then repeat the process, assuming that you are NOT allowed to repeat letters. Then subtract the two answers…</p>
<p>And I agree: this would not be an SAT question. But change the numbers to 3-letter codes from the first 5 letters and you have something that I thnk is not to hard to be the last question in a section…maybe.</p>
<p>There you are Jeffrey! I was wondering when I’d hear from you… 2416 looks right to me…</p>
<p>I just have one problem for everyone… It took me a while to figure this out.</p>
<p>Theoretically, doing (8<em>8</em>8<em>8) - (8</em>7<em>6</em>5) should yield the same result as solving the problem the long way. Can anyone explain where I made a mistake in my long solution?</p>
<p>A A _ _ 7*7 = 49 ways
A _ A _ 49 ways
A _ _ A 49 ways
_ A A _ 49 ways
_ A _ A 49 ways
_ _ A A 49 ways
A A A _ 7 ways
A A _ A 7 ways
A _ A A 7 ways
_ A A A 7 ways
A A A A 1 way</p>
<p>(49<em>6) + (7</em>3) + 1 = 323 ways for alphabet A to meet the condition.</p>
<p>323 * 8 = 2584 ways</p>
<p>There are 2584 ways to make 4 letter-codes that adhere to the condition given.</p>
<p>According to the long way, the answer is 2584. What is the mistake I made in my process?</p>
<p>When you say AA_ _ has 49 ways, included among them would be AABB, AACC etc…</p>
<p>But then you end up double counting those because when you multiply by 8 at the end, you don’t realize that you already counted some of them…did that make sense?</p>