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# some confusion?

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1) During a party, a total of 78 handshakes occurred. If each person shook hands once with each of the other people, how many people were at the party?

2) How many distinguishable ways are there to write 9 as the sum of  1's, 2's  and 4's, where the order of the addends matters? For example, 4+4+1  and 1+4+4 are two different ways.

Okay, for this one I was confused on if it had to be 3 numbers or could it be all of it? like could it be 9 ones or could it only be 3 number combos.

Feb 18, 2018
edited by Echotastic  Feb 18, 2018

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1)  One way is to consider that each person has a total of n-1 handshakes, and there are n people, so there would be (n-1)(n) handshakes, but this includes each handshake twice (1 and 2, 2 and 1) so dividing by 2 gives the correct answer. So, use the formula:

[n*(n - 1)] / 2 = 78, solve for n

n = 13 - people at the party.

Feb 18, 2018
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Yes, that's called the "handshake" formula.

tertre  Feb 18, 2018
edited by tertre  Feb 18, 2018
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What about 2)? I'm really confused on that one.

Feb 18, 2018
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(2) I think there are 96 ways...

Explanation: First, let us count the number of ways we can sum 9 with di erent combinations of 1, 2, and 4. We can have nine ones; seven ones and one two; ve ones and two twos; ve ones and one four; three ones and three twos; three ones, one two, and one four; one one and four twos; one one, two twos, and one four; or one one and two fours. Now, let us count the number of ways to rearrange each of these cases. Case One: Nine ones There is only 1 way to arrange these, since all of the ones are indistinguishable. Case Two: Seven ones and one two There are eight numbers total to sum. The two can be placed in any of the eight slots, after which the others are simply lled with ones. Hence, there are 8 ways to arrange these. Case Three: Five ones and two twos There are seven numbers total to sum. The rst two can be placed in any of the seven slots, and the second two can be placed in any of the remaining six. After this, we simply ll in the ones. However, because the twos are indistinguishable, we've overcounted by a factor of two. Thus, there are 76 2 = 21 ways to arrange these. Case Four: Five ones and one four There are six numbers to sum, and the four can be placed in any of the six slots, after which the other slots are lled with ones. So, there are 6 ways to arrange these. Case Five: Three ones and three twos There are six numbers to sum. Consider placing the twos and then lling in the ones. There are six slots for the rst two, ve for the second, and four for the third. However, given three distinguishable twos, there are six ways to arrange them. Thus, because the twos are indistinguishable, we have overcounted by a factor of six. Thus, there are 654 6 = 20 ways to arrange these. Case Six: Three ones, one two, and one four There are ve numbers to sum. The two can be placed in any of the ve slots, and then from there the four can be placed in any of the four slots and the other slots lled in with ones. Thus, there are 5  4 = 20 ways to arrange these. Case Seven: One one and four twos There are ve numbers to sum. The one can be placed in any of the ve slots, after which the twos are lled in. Thus, there are 5 ways to arrange these. Case Eight: One one, two twos, and one four There are four numbers to sum. The one can be placed in any of the four slots, the four placed after that in any of the remaining three, and the other two slots lled in with twos. So, there are 4  3 = 12 ways to arrange these. Case Nine: One one and two fours There are three numbers to sum. The one can be placed in any of the three slots, and then the other slots lled in with fours. Thus, there are 3 ways to arrange these. There are thus 1 + 8 + 21 + \(6 + 20 + 20 + 5 + 12 + 3 =\boxed{96} \)

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Feb 18, 2018
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....and I would have 96 typos if I tried to type an answer that long tertre.....strong work!

ElectricPavlov  Feb 18, 2018
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Thanks, EP! My hands hurt now..

tertre  Feb 18, 2018