First, what is h supposed to be?
n(n+1) is the sum of the first even numbers from 2 to n, i.e., 2+4+6+8+10+...
Example, 2+4+6+8+10 = 5(6) = 30
105 happens to be the sum of the first 14 integers, starting with 1, which derives from n(n+1)/2.
5050 is the sum of the first 100 intehers starting with 1 which, again, derives from 100(101)/2 = 5050.
Are you sure you have the right formula for what you are trying to determine?
two algebra questions which i need for Tuesday.
1. when h is 105 what is n in the following formula: n(n+1)
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2. the same as question one but this time h is 5050
3 answers
h is the number of handshakes
N is the number of people shaking hands with one another except h = n(n+1)/2. The problem you are dealing with must be the following.
How many handshakes will take place between a number of people in a room given that each person shakes the hand of every other person only once?
Lets look at a typical problem with real numbers.
If there are 15 people in a room and each person shakes hands with every other person in the room only once, how many handshakes will take place?
There are two ways of looking at this type of problem. The first involves determining the number of combinations of "n" things taken "r" at a time and the other the sum of all possible pairings of "n" things.
For the first method, what you are seeking is the number of possible combinations of 15 people taken two at a time with no combination being repeated in another order. What I mean here is that once "a" shakes the hand of "b", then that pairing has been made and cannot be repeated in the form of "b" shaking the hand of "a" at another time. Only the single pairings are counted. This then boils down to determining the number of possible combinations of "n" different things taken "r" at a time, C(n,r), which is derivable from the formula C(n,r) = [n(n-1)(n-2)-----(n-r+1)/r! = n!/[r!(n-r)!], where n! = n(n-1)(n-2)------(3)(2)(1), r! means r(r-1)(r-2) etc., and similarly for (n-r)!. So in your problem n = I15 and r = 2. Thus you have C(15,2) = [15x14x13----3x2x1]/(2x1)(13x12x11----3x2x1) = 15x14/2x1 = 210/2 = 105 handshakes.
The second method seeks the sum of all possible pairings between the n people as follows, using 5 people to illustrate the derivation. Person 1 shakes the hand of 4 other people. Person 2 shakes the hands of 3 other people already having shaken the hand of person 1. Person 3 shakes the hands of 2 other people already having shaken the hands of persons 1 and 2. Person 4 shakes the hand of person 5, also having already shaken the hands of persons 1, 2, and 3. So what do we have in the way of a total? All together, 4 + 3 + 2 +1 = 10.
What we did in fact is simply sum up the numbers from 4 to 1, or (n - 1) to 1. Now, the sum of any sequence of consecutive numbers from one on up is given by S = n(n + 1)/2. Notice that our sum starts with (n - 1). Therefore, the total number of handshakes between "n" people can be defined as S = [(n - 1)(n - 1 + 1)]/2 or (n(n - 1)/2.
Applying this to our 15 person problem, S = 15(14)/2 = 210/2 = 105 handshakes.
For 5050, n(n+1)/2 = 5050. Solve for n.
How many handshakes will take place between a number of people in a room given that each person shakes the hand of every other person only once?
Lets look at a typical problem with real numbers.
If there are 15 people in a room and each person shakes hands with every other person in the room only once, how many handshakes will take place?
There are two ways of looking at this type of problem. The first involves determining the number of combinations of "n" things taken "r" at a time and the other the sum of all possible pairings of "n" things.
For the first method, what you are seeking is the number of possible combinations of 15 people taken two at a time with no combination being repeated in another order. What I mean here is that once "a" shakes the hand of "b", then that pairing has been made and cannot be repeated in the form of "b" shaking the hand of "a" at another time. Only the single pairings are counted. This then boils down to determining the number of possible combinations of "n" different things taken "r" at a time, C(n,r), which is derivable from the formula C(n,r) = [n(n-1)(n-2)-----(n-r+1)/r! = n!/[r!(n-r)!], where n! = n(n-1)(n-2)------(3)(2)(1), r! means r(r-1)(r-2) etc., and similarly for (n-r)!. So in your problem n = I15 and r = 2. Thus you have C(15,2) = [15x14x13----3x2x1]/(2x1)(13x12x11----3x2x1) = 15x14/2x1 = 210/2 = 105 handshakes.
The second method seeks the sum of all possible pairings between the n people as follows, using 5 people to illustrate the derivation. Person 1 shakes the hand of 4 other people. Person 2 shakes the hands of 3 other people already having shaken the hand of person 1. Person 3 shakes the hands of 2 other people already having shaken the hands of persons 1 and 2. Person 4 shakes the hand of person 5, also having already shaken the hands of persons 1, 2, and 3. So what do we have in the way of a total? All together, 4 + 3 + 2 +1 = 10.
What we did in fact is simply sum up the numbers from 4 to 1, or (n - 1) to 1. Now, the sum of any sequence of consecutive numbers from one on up is given by S = n(n + 1)/2. Notice that our sum starts with (n - 1). Therefore, the total number of handshakes between "n" people can be defined as S = [(n - 1)(n - 1 + 1)]/2 or (n(n - 1)/2.
Applying this to our 15 person problem, S = 15(14)/2 = 210/2 = 105 handshakes.
For 5050, n(n+1)/2 = 5050. Solve for n.
0 answers