Strings

Let the number of people be $n$ and since every $2$ people out of $n$ people should be connected by a string, so the number of strings that would be used is the value equal to the binomial coefficient $\binom{n}{2}$ or $nC2$ . According to the problem we have ---

$\binom{n}{2} \lt 1000$

$\implies \frac{n(n-1)}{2} \lt 1000 \implies n(n-1) \lt 2000 \implies n^2-n \lt 2000 \implies n^2-n-2000 \lt 0$

Let us now assume that $n^2-n-2000=0$ and solve it using the quadratic formula $n=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ with $a=1,b=(-1),c=(-2000)$ for this equation. Then, we get ---

$n=\frac{1 \pm \sqrt{1-4.1.(-2000)}}{2\times 1} = \frac{1 \pm \sqrt{1+8000}}{2} =\frac{1 \pm \sqrt{8001}}{2}$

Since, $n$ cannot be (-ve), we take the value as $n=\frac{1 \pm \sqrt{8001}}{2}=\frac{1+89.45}{2}=\frac{90.45}{2}=\boxed{45.225}$

Since, we took the assumption that $n^2-n-1000=0$ while actually we have $n^2-n-1000 \lt 0$ , so we get $n \lt 45.225$

The value less than $45.225$ and still the max value of $n$ possible is $45$ , so answer is $=\boxed{45}$

use this formula (x(x-1))/2 < 1000 where x= number of people and (x-1)= number of strings per person. Dividing by two beacause strings from person A to person B is the same as Peron B to A.

Simply observing the pattern we can see that if there are n people then starting from the first person , the first person connects with (n - 1) people , the second person connects with (n - 2) people and so on till the last one connects with 1 people . So adding for n people we get (n - 1) + (n - 2) + (n - 3) + ....... + 2 +1 = (n - 1)(n )/2 Now for n = 45 we get the maximum value less than 1000 i.e. 990.

for n+1 number of guests, the number of strings are 1+2+3+4+... ... ... +n .

Assume n(n-1)/2 = 1000 so n = 44 (approx... but we take the integer value)

So the number of guests are 44+1=45

For there to be the maximum number of people n , each string must connect a different pair. If there were 3 people, there would be 3 strings, for 4 people there would be 6, for 5 there would be 10, and for 6 there would be 15. Note that the number of strings is the ( n -1)th triangle number. The highest triangle number that is still under 1000 is 990, which is the 44th triangle number. Therefore the number of people is 45.

*Triangle numbers are numbers that can be written out as that number of dots in rows to make an equilateral triangle.

Here the no. of string refers to the no. of pairs that one can form from the given number of people . Here the order does not matter since A being attached to B is equivalent to B being attached to A, hence it becomes a problem of combination and not permutation. The answer is as simple as " nC2 < 1000 " where n should be an integer, which comes at n = 45.

When people are connected, let n be the amount of people. Then the amount of connections is n(n+1)/2. We can set up an equation with 1,000 on the other side, and solve. We find that 45 is the smallest n that fits.

Imagine the people as the vertex of a polygon, and the strings as diagonals. The total number of strings would be:

$\dfrac{n(n-3)}{2} + n < 1000$ $n^2 - 3n + 2n - 2000<0$ $-44.2 < n < 45.2$ $\boxed{n = 45}$

It takes every two persons to be connected by a string. Therefore $n$ persons, it is $\dbinom n2$ . To find $\dbinom n2 < 1000$ , we have $\dfrac {n(n-1)}2 < 1000 \implies n = \left \lceil \sqrt {2000} \right \rceil = \boxed{45}$ , where $\lceil \cdot \rceil$ denotes the ceiling function .

Nice problem. I won't add another solution as there are plenty here but I really enjoyed taking a pen and paper approach to solving it.

Let's say the guests number will be $\large n_g$ and the max strings number $\large n_s$

we know logically that each two connected together with only one string

so if we start with one of the guests we will get 2 facts\info

1- $\large n_s=n_g-1$

2-The $\large n_s$ will be decreased by "1" with each guest we use as a pivot to create new pairs

$\large n_s)_{total}=1+2+3+5+.... +n_s$

using the formula

$\large Sum\space of\space numbers\space from\space 1\space to\space n=\frac{n}{2}\times(n+1)$

$\large \frac{n_s}{2}\times (n_s+1)<1000$

you can solve the equation or just try a few numbers between 40 and 50

you'll find out 44 is max number of strings and as follows 45 is the max number of guests

Since each of the people are connected then there will be a total of $_n C_2$ number of ways of doing this. Since there are only 1000 stirngs then we look for the largest value of $n$ such that $_n C_2 < 1000$ . By formula, $_n C_2 = \frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2} < 1000$ . Multiplying both by 2, we have $n(n-1) < 2000$

We assume to have equality. that is, $n (n-1) = 2000$ .
To find the value of $n$ , we look at the geometric mean of $n$ and $n-1$ which is $\sqrt{2000} = 44.7$ .

This means that the integers $n$ and $n-1$ are integers above and below 44.7. Thus we have $n = 45$ .

We can check that $45 \times 44 = 1980 < 2000$ . If we go higher, that is $n = 46$ we can see that $46\times 45 = 2070 > 2000$ .

A very simple equation can be used.(n-1)+(n-2)+(n-3)+(n-4)+...+(n-n)=strings where n=number of people. If we know the number of strings we can solve the number of people or vice versa. Hope people like it.

i solved by thought on using formula how many diagonal in n-gon + n (connect the people outside (as like the side of n-gon)) < strings -> i mean this \frac {n(n-3)}{2} + n < 1000 and i actually i got 45 as my answer.. is it my true or actually i did mistake (means i only got lucky for solving this)??

n(n+1)/2 where n=44 is equl to 990 strings, 45 people are presnt

Let's pretend there are N people at the play. Then, the person A will be tied to N-1 people, using N-1 strings. The person B will be tied to N-2 people, using N-2 strings. The person C will be tied to N-3 people, using N-2 strings etc, till the Nth people will be tied to N-N people. The total of strings is

(N-1) + (N-2) + (N-3) + ... + (N-N).

Which could be written as

(N+N+N+...+N) - (1+2+3+...+N).

And knowing that (N+N+N+...+N) has N numbers N, we may say that

(N+N+N+...+N) - (1+2+3+...+N) = N² - [N(N+1)/2] = (N² - N)/2,

that needs to be less than 1000 strings.

(N²-N)/2 < 1000

N²-N-2000<0

Resulting in -45 < N < 46. Therefore, there are a maximum of 45 people.

Let the total number of people be n. For 1 person, we need (n-1) strings since he/she ties himself/herself with everyone else except for their own self, therefore "n-1". Since 1 person requires n-1 strings, n people will require n X (n-1) strings. But wait, here is the catch- if you observe carefully, for every pair of people, only 1 string is required but here we have counted it twice. Therefore, to delete the extra strings we divide the obtained expression by 2 : n(n-1)/2 (this gives no. of strings required) Now, its pretty simple after this: Just solve the inequality n(n-1)/2 <1000 , by solving quadratic equation or by trial method and you will get the answer as 45.

I Discovered that in a polygon - The sum of no of sides + the sum of no of diagonals = n(n-1)/2
where n is the no of sides of the polygon. Eg in a triangle - 3(3-1)/2 = 6/2 = 3 since a triangle has no diagonals it gives the no of sides - 3 + diagonals - 0 >>> 3+0 = 3 In Hexagon - 6(5)/2 = 15 .. Hexagon has 6 sides and 9 diagonals ... 6+9 = 15.

So I just used the formula which I just discovered.. Wow.. n(n-1)/2 < 2000 and got the value as 45... I Just Discovered a Formula to solve this... and for no of diagonals + sides.

Here the number of strings are less than 1000. And the number of strings we need for n people to do the necessary connections in n C 2 so n C 2<1000 => n(n-1) <2000 So we need to check for the product of two consecutive numbers whose product is very near to 2000 and less than 2000. So the pair is 45, 44.

So n = 45

use excel to calculate the combination nC2, by increasing the value of n and using the formula COMBIN(n,2) , for n=45 we will get 990 strings and for n=46 we get 1035 strings, so 45 is the answer

If there are 'n' persons, then the number of strings needed to connect them is n (n-1)/2 Given, n (n-1)/2<100, solving for "n" gives 45..

lets think the number of people is X . a string is hold by two people only so the format of the combination will be (number of people=X)C2 . If there is is 3 people then the number of string will be 3C2=3 , if there is 4 people then the number of string will be 4C2 = 6 . as there is X number of people so the formula will be like

XC2 = X! / [ 2! ( n - 2 ) ! ] = X(X - 1 ) / 2 = 999 .

Here 999 come from the question , as it is said that number of person will be less than 1000 . from that formula we will get to value of X that are -44.201 ( which is inappropriate ) and other value is 45.201 . so the answer is 45 .