Hello,
my name is Irene De Lemos. I will be talking about what we learned on the 12th
and the 13th of September. On the 12th, Mr. Piatek started
the class off doing a quiz about Factorial Notation Quiz and solved each of the
questions on the quiz.
After we finished the quiz, we started a lesson on permutations with
repetitions and restrictions. The lesson began by getting volunteers to go to
the front of the class and become a family. A mother (Kyle), a father (Mathew), and 3
children (Wandaly, Charina, and I), the situation is, the ‘family’ is going to
the movies. The class had to figure out the amount of possible sitting arrangements. The twist is that the
parents are in inseparable. The class had to figure out the right amount of
permutations. Mr. Piatek showed answer and explained it thoroughly by using the
example 1, on page 19.
In
the example, there is a family of 7 consisting of a mom, a dad and 5 children.
They are going to take a family photo. We have to figure out the amount of
possibilities they could be arranged. Like the previous question, the parents
are inseparable. They have to stay together.
a) If the are no restrictions on
where they sit, how many possible arrangements are there ?
For a. we have to
figure out how many possible arrangements the family could sit together. By
using the permutation formula.
7!
(7-7)! =
(7-7)! Would equal 0!
And as we figured out that 0! Is
equal 1
The equation would become
7!
7! Is equal to 5040
By plugging in and solving, we
got 5040 different possible arrangements.
b) If the parents must be seated together, how many
possibilities are there?
We can not get the
right answer for this question by using the permutation formula. First you
should count the amount of entities in the question. An entity is a thing that independently exist. Since mom and dad want to be together the whole time, the
will be counted as one entity. 5
(the children) + 1 (parents) = 6 entities. The 6 entities will become 6!
You will end up with:
6! X 2!
You multiply 6! With 2! Because you need to multiply the factor
of number of items/people that was added up to become one entity.
6! X 2! = 1440
There are 1440 possible arrangements if the parents stay
together.
c) If the parents can not be seated together, how many
possibilities are there?
To get the answer for
c, you simply take the total amount of the possible arrangements and subtract
the amount of arrangements with the parents sitting together.
5040 – 1440 =
3600
There are 3600 possible sitting arrangements with the parents sitting apart.
d) if the parents and the oldest child be seated together,
how many arrangements possible?
d. You do the same steps as b. to get the answer.
Mom, dad, and oldest would become 1 entity plus the other
four children. This would equal 5!
Then multiply the factorial of the number of people that
made up that one entity.
5! x 3!
5! x 3! = 720
720 possible arrangements
For example 2 we have to find the amount of possible
arrangement with the numbers 2,4,5,6, and 8. If numbers are allowed to be
repeated.
a) how many four digits numbers can be formed?
Since repetitions are allowed, we cannot use the permutation
formula. We have to use the dash
method. We will need to have 4
dashes for the four spots of each number:
_____
. ______ .
_____ . ____ =
On each dash we can put all 5 five numbers because repeating the
numbers are allowed in the equation. It would look like this:
5 x 5 x 5 x 5 = 625
There are 625 possible arrangements if all five numbers are
allowed to make the four-digit number.
b) How many four digit numbers can be formed if the numbers
to be less than 5000 and divisible by 5?
We will still be using the dash method because repetitions
are allowed and there are restrictions. We will need to have 4 dashes for the
four spots of each number:
____ . _____ . _____ .
______ =
Since it has to be divisible by five, we have to put the 5
on the last dash
____ . _____ . _____ .
___1__ =
5
The four digit number has to less than 5000, the only oppositions
we have is to put either 2 or 4 as the first digit:
___1__ . _____ . _____ .
___1__ =
4 5
Or
__1__ . _____ . _____ .
___1__ =
2 5
On the two dashed in the middle, you can put all five digits
because repetitions are allowed in the question:
___1__ . ___5__ . ___5__ .
___1__ = 25
2
5
Or
__1__ . ___5__ . ___5__ .
___1__ = 25
4 5
Add the two answers 25 + 25 = 50
Another way is also putting the possible numbers right from
the start:
__2__ . ___5__ . ___5__ .
___1__ = 50
2/4 5
The answers would still be the same.
Part 2)
On
the next day we had another quiz on permutations with repeating objects
afterwards we learned a new lesson on permutations with case restrictions. On
the lesson we learned how to find the number of arrangements that are possible
with certain restrictions.
a) In how many 4 letter “words” are possible using the letters in SUNDAY?
In this question we can use the permutations formula. There are six letters and four to permute
6P4
6!
(6-4)! =
(6-4)! = 2!
6!
2!
=
To cancel out the denominator, we will need to cancel out
the permutations:
6 x 5 x 4 x 3 x 2!
6 x 5 x 4 x 3 = 360
There are 360 possible ways of arranging SUNDAY.
b) How many 4 letter “words” are possible using the letter
in SUNDAY if A is the third letter?
Well will have to use the dash method because there are restrictions in this questions. We will draw four dashes because we need to find a four-letter word:
____ . _____ . _____ .
______ =
On the third dash will have to be a one because A has to be
the letter that goes there:
____ . _____ . __1__ .
______ =
A
The rest will have to be the letters that are left in the
word. On the first dash, 5 letters are available for picking. On the second, 4
letters are left for picking. The last 3 are left to be picked.
___5_ .
___4__ . ___1__ . ___3___ = 60
A
There are 60 possible ways of arranging SUNDAY while putting
A as the third number.
In conclusion, this is what we have learned in class the
last two days. I hope you fully understood what I have written and I also wish
everything will help in anyway possible.
Your classmate,
Irene
De Lemos
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