Discrete Math

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Is this correct?

• Using the Principle of Inclusion-Exclusion, find the number of integers between 1 and 2000 (inclusive) that are divisible by at least one of 2, 3, 5, 7.

A = {n| 1 ≤ n ≤ 2000, 2 |n}
B = {n| 1 ≤ n ≤ 2000, 3 |n}
C = {n| 1 ≤ n ≤ 2000, 5 |n}
D = {n| 1 ≤ n ≤ 2000, 7 |n}

|A| = [2000/2] = 1000
|B| = [2000/3] = 666
|C| = [2000/5] = 400
|D| = [2000/7] = 285

Also, I'm kind of confused with how to approach this problem:

• John Sununus was once the governor of New Hampshire, and his name reminds one of the authors of a palindrome (a words which is spelt the same way forwards as backwards, such as SUNUNUS).

How many seven-letter palindromes (not necessarily real words) begin with the letter S and contain at most three letter?

Thanks for any helpful replies!

  • Discrete Math -

    You can add up the |A| , |B| etc. and then say that you have double counted the elements that are in both A and B, A and C, etc. etc. So, you subtract these. But then you have counted the elements that are in the intersection of A, B and C a total of zero times, because they are counted in

    |A| one time

    |B| one time

    |C| one time

    |Intersection of A and B| minus 1 time

    |Intersection of A and c| minus 1 time

    |Intersection of B and C| minus 1 time

    So, you need to add the number of elements in the intersection of A, B and C, and thus also all other pairs of intersections of 3 sets.

    Proceeding in this way, you will find the usual inclusion-exclusion rule. You can derive this more formally, by first computing the number of elements that are not divisible by any of the numbers.

    You can then directly apply the inclusion-exclusion formula in its usual formulation, so you then find that this is the total number of elements minus |A|, minus |B|,.. plus |A intersection B|, etc. etc. etc.

    This is then also N minus the number of elements divisible by either of the numbers, you get the same result.

  • Discrete Math -

    Okay I continued the first problem:

    |A ∩ B| = [2000/6] = 333
    |B ∩ C| = [2000/15] = 133
    |C ∩ D| = [2000/35] = 57
    |A ∩ D| = [2000/14] = 142
    |A ∩ B ∩ C ∩ D| = [2000/210] = 9

    1000 + 666 + 400 + 285 - 333 - 133 - 57 - 142 + 9 = 1695 <--Answer

  • Discrete Math -

    I too am having trouble with that SUNUNUS problem, any idea how to approach it?

  • Discrete Math -

    Any suggestions?

  • Discrete Math -

    The only thing I can think of is starting with the method from the first problem and plugging in the numbers? How have you started it?

  • Discrete Math -

    To be honest I haven't started yet, but your method sounds like a step in the right direction. . .I'll play around with it for a little and see what I get. . .If you figure out anything post. Hopefully someone who knows something will post cuz I'm lost

  • Discrete Math -

    Yeah same here and it's due tomorrow lol. So far I got this for #2 but I think it's wrong.

    A.) 1 x 26 x 25 x 24 = 15,600
    B.) 26 x 25 x 24 x 23 = 14,398
    C.) 25 x 24 x 23 = 13,800

  • Discrete Math -

    Francesca,

    INCLUSION/EXCLUSION PRINCIPLE:
    Your post: 2011-04-12T15:08

    I believe the present problem can be solved using the principle to calculate the count of numbers NOT divisible by ANY of the 4 factors (2,3,5,7). Subtract that from 2000 to get the count of numbers divisible by at least one of the four factors.

    The inclusion/exclusion principle works as follows:
    For a two set case, and using u to denote the cardinality of the universal set (2000),
    a=count of numbers divisible by 2,
    b=count of numbers divisible by 3, then
    ab=count of numbers not divisible by 2*3 (i.e. |A∩B|)

    Count of numbers NOT divisible by either factor (2 or 3) is:
    N̅=u-(a+b)+ab
    =2000-(1000+666)+333
    =667

    For the case of 3 factors,
    N̅=u-(a+b+c)+(ab+bc+ca)-(abc)

    For the case of 4 factors:
    N̅=u - (a+b+c+d) + (ab+ac+ad+bc+bd+cd) - (abc+abd+bcd+acd) +abcd
    where
    a=count of numbers divisible by 2
    ab=count of numbers divisible by 2*3
    abc=count of numbers divisible by 2*3*5
    abcd=count of numbers divisible by 2*3*5*7

    If you proceed this way, you should get the count of numbers NOT divisible by any of the 4 factors as:
    N̅=2000-2351+960-160+9=458

    So the count of numbers divisible by AT LEAST one of the four factors 2,3,5,7 is 2000-458=1542.

    If you have questions, please post.

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