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So much counting!

  1. Jun 12, 2006 #1
    1.The integer 388,800 can be factored with primes as 2^6 × 3^5 × 5^2
    (a) How many unique divisors does 388,800 have?
    (b) How many of these factors are even? odd?

    I have no clue how to do this. There are no similar examples in our textbooks or notes. I searched up counting divisors on the internet and couldn't understand it very well. Help?

    2.The letters a, b, c, d, e, f, g can be arranged to form a number of words or sequences of up to length 7 [no repeated letters are allowed].
    (a) Let S be the set of all words of length 7. What is n(S)?
    (b) Let X be the set of all words of length 6. What is n(X)?
    (c) Let U be the set of all words from length 1 to length 7. What is n(U)?
    (d) How many words in U do not begin with a?

    Do I solve a) and b) with permutations? Why is it that both of them yield the same answer?

    n(S) = 7!/0 = 7!
    n(X) = 7!/ (7-6)! = 7!

    What would I do for c) and d)?

    3. A and B are sets where n(U) = 500, n(A) = 223, n(B) = 333, and n(A()B) = 133. Determine:
    (a) n(A()B′), () = the intersection of

    If I know n(A) = 223 and n(B') = 500-333=167, it doesn't seem plausible to find how many elements are in both (but I know there is a way!).
    Last edited: Jun 12, 2006
  2. jcsd
  3. Jun 12, 2006 #2
    You can try this.
    You have 3 set

    A={2,4,8..} n(A)=6
    B={3,9,27,.} n(B)=5
    C={5,25} n(C)=2

    the total number of divisors are

    n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)

    where AXB are number formed by multipliying each element of A for each element of B, and taking into account
    that aixbi = bixai

    for instance n(AXB)=30

    for b) you just have to remember that what happens when you multiply, odd*odd, even*even, even*odd
  4. Jun 13, 2006 #3
    Hey thank you, that was neat!

    Anyone have suggestions for 2) and 3)?

    Also, I'm having trouble proving these two statements:
    http://img115.imageshack.us/img115/8898/proof1tz.jpg [Broken]

    For a), I have:

    LS: n!/(n-4)!4! - (n-1)!/(n-5)!4!
    = n(n-2)(n-3)/4! - (n-1)(n-2)(n-3)(n-4)/4!

    RS: (n-1)!/(n-4)!3!
    = (n-1)(n-2)(n-3)/3!

    Then I got stuck...

    I had even more difficulty with b)
    Last edited by a moderator: May 2, 2017
  5. Jun 13, 2006 #4
    For question 3, it may be helpful for you to note that [tex]n(A \cup B)=n(A)+n(B)-n(A \cap B)[/tex]. Also, what conclusion can you make when [tex]n(U)[/tex] is greater than [tex]n(A \cup B)[/tex]?

    For part a of the last question you posted, your approach is correct but you seem to have left out an (n-1) in one of your expressions. Apart from this, all you'll need to do is to carry out some factorization and your answer should appear quite quickly! Factorization is also an important step in solving part b!
    Last edited: Jun 13, 2006
  6. Jun 13, 2006 #5
    1)Let's say a no X is written in terms of its prime factors like
    [tex]X={a^m} \times {b^n}\times ...... [/tex]
    Then the total no. of possible factors for X ( including ! and X itself)
    is given by (m+1)(n+1) .....
    I'll leave the proof to you as an exercise :)
    Hint :Think principle of independent counting .
    Once you figure out the proof, then the second question should be a piece of cake.

    Well, why not ?
    The greater no. of ways of arranging in the first case is compensated in the second case by a greater no. of ways of choosing .
    To make it clearer,
    n(S) = 1*7!
    n(X) = 7*6!
    You can easily see the "compensation" .

    c) Can be done, perhaps by simply adding no. of words of each length together ie, No of 1 letter words + No of 2 letter words + ......
    It's a bit tedious nevertheless.
    d)In how many ways can you choose the first letter ?
    Multiply that with the no. of ways in which you can arrange the remaining letters .

    3)Drawing a Venn diagram makes it very simple.
    Also, I believe there are other parts to the question, as information given seems excessive.
    Last edited: Jun 14, 2006
  7. Jun 13, 2006 #6

    matt grime

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    you just do it; count them. (you understand what unique factorization implies, right.....)
  8. Jun 14, 2006 #7
    i still cant get part b)

    on the right side, after the equating the first line, i get

    (n-1)!/(n-r)!(r-1)! + (n-1)!/(n-r-1)!r!

    what can i do to simplify that?
  9. Jun 14, 2006 #8
    Consider, r! = r (r-1)!

    Can you work out a similar expression for (n-r)! and (n-r-1)! ?

    With this, you should be able to carry out some factorization to obtain your final answer.
  10. Jun 14, 2006 #9
    Arrrrg, I'm still having an immense amount of difficulty with this stuff. pizzasky, i got part b) of that question but the original questions I posted, I'm still struggling with.

    i got an answer for number 1 from my original post, but could you please spell out the proof?

    for 2.c), are you sure that method is right? are you suggesting i do 7!/(7-1)! + 7!/(7-2)! +...+ 7!/(7-7)!?

    for 3, the venn diagram didnt help. even though i know how many are in A and how many are in B', how do I know which ones are common among these two?
    Last edited: Jun 14, 2006
  11. Jun 14, 2006 #10
    For Q3, how is [tex]n(A)[/tex] related to [tex]n(A \cap B)[/tex] and [tex]n(A \cap B')[/tex]?

    For Q2c, I agree with the method arunbg suggested, but feel free to clarify your doubts/ concerns.

    For Q1, if you can understand the method suggested by mathphys, you should be able to understand arunbg's approach to the question! (But mathphys's working seems to have left out the factor 1. The approach by arunbg has taken this into account)
  12. Jun 14, 2006 #11
    with regards to Q2, my apologies, i didnt consider that ab and bc were different entitites.

    however, can you take a look at this question:

    Kathryn is interested in purchasing a new automobile. The basic car, comes with power steering, power disc brakes, automatic transmission and an AM/FM radio. The brochure shows 10 additional options available, at extra cost, of course, for the car she intends to purchase.

    The option list is given as follows:
    AM/FM radio with CD player, sunroof, leather interior seats, air conditioning, cruise control, interval wipers, sport wheel covers, custom floor mats, chrome body moulding, remote side mirrors

    (a) How many different configurations of options can Kathryn choose?
    (b) Kathryn has decided to get the AM/FM radio with CD player and the sunroof. How many configurations are now possible?

    so for this question, something like sunroof and cruise control versus cruise control and sunroof would be considered the same thing. now how do i approach this?
  13. Jun 15, 2006 #12
    Last question, use the fact that you can either choose or not choose a particular option ie, you have two options with each accessory independently.The overall configuration is changed by either choosing or not choosing specific options. Can you go from here ?

    Earlier questions
    1) Proof Hint :
    The method is similar to mathphys as pizzassky pointed out.
    As an eg, 72 = 2^3 * 3^2 = 1*(2^3) + 1*(3^2)
    There are three different multiples of 2 (2,4,8) and two of 3 (3,9) in terms of which the no. can be written. A combination of any multiple of 2 and 3 leads to a factor. Also there can be multiples of 2 alone and 3 alone as factors. So how many combinations are possible ?

    2)I don't see any other method .

    3) Write [tex]n(A \cap B')[/tex] in terms of [tex]n(A)[/tex] and [tex]n(A \cap B)[/tex]
  14. Jun 15, 2006 #13


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    A much simpler approach would be to calculate 7*6*3-1, where it should be obvious how we get the numbers 7, 6 and 3.

    We get your result, of course, by expanding (6+1)*(5+1)*(2+1)-1
    Last edited: Jun 15, 2006
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