third open query

course Mth 151

your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

003. `query 3

question 11.3.20 5 prizes among 25 students

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

25*24*23*22*21=25!/((25-5)!)

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:: ** There are 25 students available so there are 25 choices for the first student.

On the second choice there are 24 students left so there would be 24 possibilities.

Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices, respectively.

The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices.

25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since

25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21

25 ! / ( (25 - 5) !) is P(25, 5).

We use permutations because in this case, there are 5 different prizes so the order in which the students are chosen makes a difference in the final outcome. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

*********************************************

question: Is repetition allowed in this situation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution: no, because repetition would mean that more than one person recieved one or more prizes, which is unfesible for the equation. 5 seperate prizes go out to 5 seperate people.

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:: ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

question 11.3.30 3-letter monogram all letters different

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

if the alphabet contains 26 seperate letters there will be 26 choices for the first piece,

25 for the second, and 24 for the third. so let us use the fundamental counting principle, wherein P(26,3)= 26!/(26-3) ! = 26*25*24. For this problem however, the initials could not repeat because the third initial must remain the same. Hence one initial (lettered tile) has already been chosen, so the equation drops one factorial place. becoming 25*24=600 possibilities.

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:: ** We are choosing 3 different letters, and since the monogram will be different if you change the order of the letters, we can say that order definitely applies.

If there is no restriction on any letter, other than the restriction of no repetitions, then there are 26 choices for the first letter, 25 for the second, 24 for the third so by the Fundamental Counting Principle there are 26 * 25 * 24 ordered choices.

We can write this as P(26, 3), the number of possible permutations of 3 objects chosen without replacement from 26 possible objects.

P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression.

However in this question the third initial must be the same as Judy's, which is `z'. Thus, since there can be no repetitions, there are only 25 possibilities for the first letter (can't be `z') and 24 for the second (can't be `z', can't be the first).

So there are only 25 * 24 = 600 possibilities. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

question 11.3.43 /& 42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

25-3=22 22-4=18 18-5=13 13-6=7 7-7=0

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:: ** We can choose the groups in any order we wish. Each group chosen is chosen without regard for order.

If we choose to begin by making the group of 3, there are 25 students available when we begin to select our group, so there are C(25, 3) possible choices.

If we make the group of 4 next there are 22 students left from whom to choose so there are C(22, 4) possible choices.

If we make the group of 5 next there are 18 students left from whom to choose so there are C(18, 5) possible choices.

If we make the group of 6 next there are 13 students left from whom to choose so there are C(13, 6) possible choices.

If we make the group of 7 next there are 7 students left from whom to choose so there are C(7, 7) possible choices.

The Fundamental Counting Principle says you that you have to multiply the number of ways of obtaining the first group by the number of ways of obtaining the second group by the number of ways of obtaining the 3rd group by the number of ways of obtaining the fourth group by the number of ways of obtaining the fifth group. So get have

C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) ways to complete the grouping.

Note that we could have chosen the groups in a different order, perhaps with the group of 7 first, the group of 6 second, etc.. The same reasoning would tell us that there are now C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) ways to do complete the groupings.

The question is, would this make a difference in the final result?

To find out we compare the two results C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3).

If the two expressions C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) are both written down and simplified, both turn out to have exactly the same numbers in their numerators, and the same numbers in their denominators. As a result, they both end up in the same form

25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ].

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

question 11.3.60. C(n,0)What is the value of C(n,0)? What is the value of C(8,0)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

anything with 0 in the r spot = 0

.............................................

Given Solution:

** C(n, r) is the number of ways of choosing r objects out of n available objects, without regard for order.

C(n, 0) is therefore the number of ways to choose 0 objects from among n objects.

No matter what n is, there is exactly one way to do this, which is to choose nothing.

Thus C(n, 0) is always equal to zero.

As another example:

There are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads.

Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) is the number of ways to obtain 0 Tails. So C(4,0) must be 1.

The formula also gives us the same result:

C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. ** "

You've submitted the assignment and you'll get credit for having done so.

Do note that in order for me to give you useful feedback, you need to include confidence assessments (which are missing in this document) and follow the self-critique process (which you have used in some, but not all, of the places where it's appropriate.

It isn't required, but if you feel that more feedback would be helpful you are welcome to add this information and submit this assignment again.

&#Let me know if you have questions. &#