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Sav [38]
3 years ago
14

In a necklace, there are 2 white beads for every 7 purple beads. There is a total of 8 white beads. How many purple beads are in

the necklace?
Mathematics
1 answer:
tresset_1 [31]3 years ago
4 0

Answer:

OK! So this is a case of circular permutation. While dealing with this type of problems, we consider the circle to be placed in free space i.e it is symmetric when viewed from and point around it. So to break the symmetry we first fix one of the beads.

Here I am taking a red bead. Now we are left with 4 red and 2 blue beads.

6 beads can be placed in 6! Ways as 1st place will have 6 possibilities, 2nd place will have 5 possibilities and so on which gives

6*5*4*3*2*1 = 6!

But this time we have considered some extra cases as well because the red and blue beads are identical so which red bead has occupied a specific position will not at all matter. So we divide the total by 5! For identical red beads (the fixed red bead will also affect the position of other beads) and 2! For identical blue beads.

Finally we have 6!/(5! *2!) = 720/(120*2) = 3.

So we can have 3 different necklaces.

As the number is small, we can even enlist the possibilities…

B B R R R R R

B R B R R R R

B R R B R R R

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The spaces are put in to help make the number more readable. Note how the "153846" keeps repeating forever

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24. A political discussion group consists of 30 Republicans,
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Answer:

12 /67

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3 years ago
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What is the bigger number, a googol or a billion?
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4 years ago
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In how many ways can a quality-control engineer select a sample of3 transistors for testing frm a batch of 100 transistors?
Advocard [28]

Answer:

161700 ways.

Step-by-step explanation:

The order in which the transistors are chosen is not important. This means that we use the combinations formula to solve this question.

Combinations formula:

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C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

3 transistors from a set of 100. So

C_{100,3} = \frac{100!}{3!(100-3)!} = 161700

So 161700 ways.

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3 years ago
Determine the mean, median, mode, and range for the data set:
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