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Daniel [21]
3 years ago
13

I have a circular necklace with $18$ beads on it. All the beads are different. Making two cuts with a pair of scissors, I can di

vide the necklace into two strings of beads. If I want each string to have at least $6$ beads, how many different pairs of strings can I make?
Mathematics
1 answer:
algol133 years ago
3 0

Solution:

There are four general types we can make: (6,12),(7,11),(8,10),(9,9).

First type: (6,12), there are 18 possible ways to choose those 6, which are going to be cut from the necklace: choose a direction to count the beads, and choose a starting position (between 2 beads). There are exactly 18 starting positions, since there are 18 spaces between the beads. One cut is equivalent for a pair we can make.

Second type: (7,11), with the same reasoning, there are 18 possible ways to cut the necklace.

Third type: (8,10), with the same reasoning, there are 18 possible ways to cut the necklace.

Fourth type: (9,9), The same reasoning cannot be applied again, since half of the cuts would be exactly the same as the other half. So there are 9 possible cuts, exactly one or each axis of symmetry for the necklace.

The solution will be the sum these values: 18+18+18+9=63

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Answer:

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Step-by-step explanation:

Given:-

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                           M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\

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                           M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\

                           M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] }  \, dtheta\\\\M = 5742\pi  kg              

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