If 190 pairings possible, how many are in group?
- Thread starter mcheytan
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What formulas have they given you for this sort of exercise? Or, if they haven't provided you with any formulas or algorithms, what (smaller) numbers have you worked with, to try to find a pattern, and what were the results?mcheytan said:
Please be complete. Thank you!
Eliz.
Re: Simple word problem
Hello, mcheytan!
This is a "combinations" problem.
If you're familiar with the formula, it's easy . . . well, sort of . . .
\(\displaystyle \text{We have: }\;{n\choose2} \:=\:190\quad\Rightarrow\quad \frac{n!}{2!(n-2)!} \:=\:190 \quad\Rightarrow\quad \frac{n(n-1)}{2} \:=\:190\quad\Rightarrow\quad n(n-1) \:=\:380\)
\(\displaystyle \text{We have a quadratic equation: }\;n^2-n-380 \:=\:0\)
. . \(\displaystyle \text{which factors: }\;(n-20)(n+19) \:=\:0\)
. . \(\displaystyle \text{and has roots: }\;n \;=\;20,\,-19\)
\(\displaystyle \text{Therefore, there are }20\text{ people in the group.}\)
Hello, mcheytan!
This is a "combinations" problem.
If you're familiar with the formula, it's easy . . . well, sort of . . .
\(\displaystyle \text{We have: }\;{n\choose2} \:=\:190\quad\Rightarrow\quad \frac{n!}{2!(n-2)!} \:=\:190 \quad\Rightarrow\quad \frac{n(n-1)}{2} \:=\:190\quad\Rightarrow\quad n(n-1) \:=\:380\)
\(\displaystyle \text{We have a quadratic equation: }\;n^2-n-380 \:=\:0\)
. . \(\displaystyle \text{which factors: }\;(n-20)(n+19) \:=\:0\)
. . \(\displaystyle \text{and has roots: }\;n \;=\;20,\,-19\)
\(\displaystyle \text{Therefore, there are }20\text{ people in the group.}\)
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Well, never mind now.... :roll:stapel said:
But for future questions of this sort (such as on tests, where there won't be somebody to complete the exercises for you), you'll need to understand the process:
Start with small numbers, and try to find a formula. If you have two, A and B, in the crowd, you have one pair: AB. If you have three, A, B, and C, in the crowd, you have three pairs: AB, AC, BC. And so forth. At each stage, by adding one more person to the group, you have all the original pairings (from the last step), plus an addition number of pairs, equal to the number of people in the previous step, because of each previous member being paired with the new guy. That is, at any stage, the number of pairings will be the sum of (the number of people in the group before this stage) and (the number of pairings at the previous stage).
So you have:
. . . . .n = 2: 1 pair
. . . . .n = 3: 2 + 1 = 3 pairs
. . . . .n = 4: 3 + 3 = 6 pairs
. . . . .n = 5: 4 + 6 = 10
...and so forth.
You can continue on, until you find the value of "n" which returns a value of "190" pairings, or you can try to generate a formula.
Hope that helps!
Eliz.