Yahoo Answers: Answers and Comments for Please answer this Urgent? [Mathematics]
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From andyjalc
enAU
Thu, 12 Feb 2009 19:46:51 +0000
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Yahoo Answers: Answers and Comments for Please answer this Urgent? [Mathematics]
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Thu, 06 Oct 2016 02:06:50 +0000
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From think613: These are all basic permutation problems. To f...
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Thu, 12 Feb 2009 20:04:33 +0000
These are all basic permutation problems. To figure out how many combinations can be made, you would multiply the amount of choices for the first place by the amount of options for the second place, then multiply that by the amount of choices for the third place, and so on.
But be careful  sometimes using one for the first place means that you can't use it again for the second place, as in the first three options.
You can also do this visually, by making a tree diagram: draw all your options for the first place. Then by each one draw what can go after it. Etc.
For example, let's do #1:
The scrabble tray has seven letters, and each can be used only once. You want a fourletter arrangement, which means there are four positions to be filled. So you would do 7x6x5x4, because there are seven choices for the first letter, and once one of those is used you only have 6 options for the second, and so on. (in contrast to the license plate problem, in which you have the same amount of choices for each letter space because you can use a letter more than once. So that would be 26x26 for the letters, then x9x9x9 for the three digits)
The visual method would look like this:
FERS
FERX
FERA
FERI
FESR
FESX
FESA
FESI
...and so on. This could take forever, but it helps to do it once or twice so the math version makes sense.

From Pi R Squared: Hi,
1. A scrabble tray contains the tiles F...
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Thu, 12 Feb 2009 20:03:11 +0000
Hi,
1. A scrabble tray contains the tiles FERSXAI. How many different fourletter arrangements can be made?
Choosing 4 of 7 letters to make different fourletter arrangements is a permutation of 7 letters taken 4 at a time, where order is important because it would make different arrangements.
7nPr4 = 840 <==ANSWER
2. In how many ways can a committee of two boys and three girls be formed from a group of 10 boys and 12 girls?
This is a combination of 2 of 10 boys multiplied by a combination of 3 of 12 girls.
10nCr2 * 12nCr3 = 45 * 220 = 9,900 <==ANSWER
3. How many different ways can three chocolate, four strawberry, and two butterscotch sundaes be served to nine people?
Nine things could be handed out 9! different ways, but when there are repeated items, 9! must be divided by the factorials of how many times each repeated item occurs. For this problem, it is:
. .9!
 = 1,260 <==ANSWER
3! 4! 2!
4. An auto license plate is made using two letters followed by three digits. How many license plates are possible?
26 * 26 * 10 * 10 * 10 =676,000 <==ANSWER
I hope that helps!! :)

From Andy J: 1. Order is important, so we use permutations....
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Thu, 12 Feb 2009 19:55:05 +0000
1. Order is important, so we use permutations. Since there are no repeated letters, the answer is simply 7P4 = 840
2. 10C2 * 12C3 = 45 * 220 = 9900
3. Assuming every person gets exactly one sundae. We simply need to find the permutation of the 9 sundaes. Normally this is 9!, but with repeats, we need to divide out the indistinguishable permutations of the 3, 4 and 2 identical sundaes. The answer is thus:
9! / (3!4!2!) = 1260
4. 26 * 26 * 10 * 10 * 10 = 676000