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2 years ago

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HURRY I NEED HELP NOW PLS. For his lunch, David is making a sandwich that must consist of bread, cheese, and meat. David can cho

ose from white, French or rye bread, either American or Swiss cheese, and the choice of turkey, ham or pastrami as a meat. Create a tree diagram to represent the combinations of bread, cheese, and meat David could make for his sandwich

1 answer:

Tom [10]2 years ago

8 0

3 types of bread, 2 types of cheese, and 3 types of meat.

I have attempted a tree diagram below:

A—> american cheese

S—>Swiss cheese

T—>turkey

P—>pastrami

H—>ham

white bread french bread rye bread

| | |——————- —————- -—————

| | | | | | | | |

T H. P. T. H. P. T. H. P

| | | | | |. | | |

—- —- —- —- —- —- —- —- —-

| | | | | | | | | | | | | | | | | |

A S A. S. A. S A. S. A. S A. S. A. S. A. S A S

Therefore you can know the number of combinations by counting the branches at the bottom so there are 18 combinations total. I hope this helps

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2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

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The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

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