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zysi [14]
3 years ago
13

Consider the problem of distributing 10 distinct books among three different people with each person getting at least one book.

Explain why the following solution strategy is wrong: first select a book to give to the first person in 10 ways; then select a book to give to the second person in nine ways; then select a book to give to the third person in eight ways; and finally distribute the remaining seven books in 73 ways.
Mathematics
1 answer:
kakasveta [241]3 years ago
6 0

The strategy used is not well, because it becomes very complex and very difficult to calculate the ways, it would play one by one, that is, despite making sense, the last part of the way to distribute the books how is it obtained? why would it be like this?

There is a much more effective way of solving the problem, and that is to divide it by cases.

Case 1:

Choose four books from the 10 available books and give them to the first student

Then choose three books from the remaining six books and give them to the second student.

Finally, give the remaining three books to the last student

Case 2:

Choose three books from the 10 available books and give them to the first student

Then choose four books from the remaining seven books and give them to the second student.

Finally, give the remaining three books to the last student

Case 3:

Choose three books from the 10 available books and give them to the first student

Then choose three books from the remaining seven books and give them to the second student.

Lastly, give the remaining four books to the last student

First it is to calculate the number number in each case by means of combinations, multiply the options per case and then add the final value in each case.

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Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132
Ilia_Sergeevich [38]

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

The HCF is the product of the highest factors

So, the HCF is:

HCF = 2 * 3 * 11

HCF = 66

<h3>How to determine the LCM</h3>

In (a), we have:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

So, the LCM is:

LCM = 2^3 * 3 * 7 * 11

LCM  = 1848

Hence, the HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

Read more about prime factorization at:

brainly.com/question/9523814

4 0
1 year ago
Solve 2tan(x)cos^2(x)=1 algebraically
deff fn [24]
X= pi/4 +k(pi)
(This may not be correct sorry)
7 0
1 year ago
Write an equation for the quadratic graphed below
uranmaximum [27]

Answer:

Step-by-step explanation:

LOL The graph doesn’t match the y intercept :)

Anyway

If we have point (0,-3) we have a quadratic of

y=ax^2+bx-3 we are given points (-1,0) and (2,0) so

a-b-3=0 and 4a+2b-3=0

4a+2b-3+2(a-b-3)=0

4a+2b-3+2a-2b-6=0

6a-9=0

6a=9

a=1.5, since a-b=3

1.5-b=3

b=-1.5

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6 0
3 years ago
Read 2 more answers
What is the 214 term in the sequence 7,10,13,16?
mixer [17]
We can figure this out using the explicit formula.

f(n)=f(1)+d(n-1)

n represents the term we are looking for.
f(1) represents the first term in the sequence, which in this case, is 7.
d represents the common difference, which in this case, is +3.

f(n) = 7 + 3(n - 1)
f(n) = 7 + 3n - 3
f(n) = 4 + 3n

Now, we can input 214 for n and solve.

f(214) = 4 + 3(214)
f(214) = 4 + 642
f(214) = 646

The 214th term in this sequence is 646.
5 0
3 years ago
The account balance of each of the three children at the end of the month is shown below
Diano4ka-milaya [45]
Just plot the values onto a normal number line

6 0
3 years ago
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