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DIA [1.3K]
2 years ago
6

I run a book club with n people, not including myself. Every day, for 365 days, I invite three members in the club to review a b

ook. What is the smallest positive integer n so that I can avoid ever having the exact same group of three members over all 365 days?
Mathematics
1 answer:
Bezzdna [24]2 years ago
7 0
<h3>Answer:   15</h3>

========================================================

Explanation:

The order doesn't matter. A group like {A,B,C} is the same as {B,A,C}.

All that matters is the overall group rather than the positioning of the members.

We'll use the nCr combination formula since order doesn't matter.

The value of n is unknown, but we know that r = 3 members are to be selected.

Let's pick a value for n at random. Let's say n = 10.

Plug n = 10 and r = 3 into the nCr formula below.

n C r = \frac{n!}{r!(n-r)!}\\\\10 C 3 = \frac{10!}{3!*(10-3)!}\\\\10 C 3 = \frac{10!}{3!*7!}\\\\10 C 3 = \frac{10*9*8*7!}{3!*7!}\\\\ 10 C 3 = \frac{10*9*8}{3!}\\\\ 10 C 3 = \frac{10*9*8}{3*2*1}\\\\ 10 C 3 = \frac{720}{6}\\\\ 10 C 3 = 120\\\\

Unfortunately we don't reach 365 or larger.

Let's try n = 11

n C r = \frac{n!}{r!(n-r)!}\\\\11 C 3 = \frac{11!}{3!*(11-3)!}\\\\11 C 3 = \frac{11!}{3!*8!}\\\\11 C 3 = \frac{11*10*9*8!}{3!*8!}\\\\ 11 C 3 = \frac{11*10*9}{3!}\\\\ 11 C 3 = \frac{11*10*9}{3*2*1}\\\\ 11 C 3 = \frac{990}{6}\\\\ 11 C 3 = 165\\\\

We're still under our target. The good news is that the nCr value is increasing.

So the idea is to do trial and error with various values of n. Keep incrementing n until nCr = nC3 is equal to 365 or larger.

Here's a table of values where r = 3 the entire time

\begin{array}{|c|c|} \cline{1-2}\text{n} & \text{nCr}\\\cline{1-2}10 & 120\\\cline{1-2}11 & 165\\\cline{1-2}12 & 220\\\cline{1-2}13 & 286\\\cline{1-2}14 & 364\\\cline{1-2}15 & 455\\\cline{1-2}\end{array}

The nCr values are also found in Pascal's Triangle. Each of those values are the fourth entry of each row.

When n = 14, we have nCr = 364 which is very close. We're one short unfortunately.

So we have to go for <u>n = 15</u> instead. This makes the nCr value well over 365 of course, but it guarantees that you'll have plenty of trios to choose from such that no group of three is repeated. Unfortunately some trios will be left out.

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Salsk061 [2.6K]
The correct answers are C, D, and E.

1 quart = 32 fl oz
1 pt = 16 oz

Using these conversions you will multiply the number of quarts or pints by the number of ounces in each and then add the totals together to find if they equal 128 oz.

C.   4 x 32 = 128 oz
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8 0
3 years ago
N/2 + 0.6 =2<br> PLS HELP
iogann1982 [59]

First, subtract 0.6 from both sides of the numbers and then simplify

Then multiply all terms by the same value to eliminate fraction denominators which gives us 2.8

So, n = 2.8

Hope it helps!

6 0
3 years ago
if there are 90 calories in 3/4 cuo of yogurt how man calories are in 3 cups of yogurt 30 calories 202 calories 270 calories 360
Debora [2.8K]
Hello!

To find how many calories in three cups we can use the proportion below.

\frac{calories}{cups} = \frac{90}{3/4} = \frac{c}{3}

To find our answer we can say, "How many times does 3/4 go into 3," and then multiply that number by 90 as shown below.

3÷3/4=4
90(4)=360

Just to verify, we will cross multiply and check our answer 

90(3)÷3/4
270÷3/4=360

Therefore, our answer is D) 360 calories.

I hope this helps!

6 0
3 years ago
Read 2 more answers
Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (
IrinaVladis [17]

Answer:

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14  and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median  of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = \dfrac{9+9}{2}

median = \dfrac{18}{2}

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = \dfrac{12+13}{2}

Median = \dfrac{25}{2}

Median = 12.5

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = Q3 - Q1

The interquartile range =  12.5 - 9

The interquartile range = 3.5

7 0
3 years ago
G(5)= please help asap
xenn [34]

Answer:

The answer is 5G

Step-by-step explanation:

6 0
3 years ago
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