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Math Help - Equation problem and others...

  1. #1
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    Equation problem and others...

    Could any1 help me a bit with this equation ?
    1) X1+X2+X3=11 , So how many are the solutions?

    2) We 've got these letters -> A,B,C,D,E,F,G,H , how many permutations
    of these letters can we make that include ABC ?

    3) How many ways are possible in order 3 girls and 8 boys sit in a row so that Mary and John NOT to sit next to each other ?

    4) I can understand at all what this question wants, but anyway ill ask here maybe you can understand ...

    How many bytes exist with 1) Exactly 2 aces
    2) Exactly 4 aces
    3) Exactly 6 aces
    4) At least 6 aces

    i dont get it , a byte consists of 8 bits, whats that ace? any ideas?

    Thanks in advance for your help =)
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  2. #2
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    Hi primeimplicant,
    where are you getting stuck with these?

    For 1)

    X_1,\ X_2, X_3 probably need to be positive integers >0.

    Are any of the numbers allowed to be zero?

    If they are allowed to be negative or non-integers, that leaves infinitely
    many solutions.
    -14+1+2, -15+1+3 etc

    Hence, start with 1, add 2 and find out what X_3 is.
    Then start again at 1, add 3 and find out what X_3 is.

    When finished with 1, go on to 2 and make sure you leave out all numbers
    less than 2.
    Then go on to 3, leaving out all numbers less than 3.
    Continue until done, there are only a few.
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  3. #3
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    Quote Originally Posted by primeimplicant View Post
    [snip]
    4) I can understand at all what this question wants, but anyway ill ask here maybe you can understand ...

    How many bytes exist with 1) Exactly 2 aces
    2) Exactly 4 aces
    3) Exactly 6 aces
    4) At least 6 aces

    i dont get it , a byte consists of 8 bits, whats that ace? any ideas?

    Thanks in advance for your help =)
    My guess is that an "ace" is a 1 bit, although this is unusual terminology I have not seen before. If my guess is correct then 1) is asking you how many 8-bit sequences there are which consist of two 1's and six 0's.
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  4. #4
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    Hello, primeimplicant!

    1)\;x_1+x_2+x_3\:=\:11
    So how many are the solutions?
    The problem is NOT clearly stated.
    I must assume that the x's are positive integers.

    Consider an 11-inch board marked in 1-inch intervals.

    . . \square\square\square\square\square\square<br />
\square\square\square\square\square


    We will cut the board at two of the ten marks.

    Then: . \square\square\square\square\square |<br />
\square\square | \square\square\square\square .represents: . 5+2+4

    And:- - \square \square \square | \square \square \square \square \square \square \square | \square .represents: . 3 + 7 + 1


    Therefore, there are: . _{10}C_2 \:=\:45 solutions.



    2) We 've got these letters: . A,B,C,D,E,F,G,H
    How many permutations of these letters can we make that include ABC ?
    I assume you mean in that exact order: ABC

    Duct-tape ABC together.

    Then we have 6 "letters" to arrange: . \boxed{ABC}\;D\;E\;F\;G\;H

    There are: . 6! \,=\,720 permutations.



    3) How many ways are possible in order 3 girls and 8 boys sit in a row
    so that Mary and John do NOT sit next to each other?
    With no restrictions, the children can be seated in: 11! ways.


    Suppose Mary and John DO sit together.
    Duct-tape them together.

    Then we have 10 "people" to arrange: . \boxed{MJ}\;A\;B\;C\;D\;E\;F\;G\;H\;I
    . . They can be seated in: 10! ways.

    But Mary and John could be taped like this: .  \boxed{JM}
    . . This makes for another 10! ways.

    So there are: 2(10!) ways that Mary and John DO sit together.


    Therefore, there are: .  11! - 2(10!) \:=\:39,916,800 - 2(3,628,800) \:=\:32,659,200 ways
    . . that Mary and John do NOT sit together.

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  5. #5
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    Thanks a lot Soroban , awesome, i totally understood!!
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  6. #6
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    Hi Primeimplicant,

    For Q1...

    that was fabulous work by Soroban!!
    He's an artist at breaking long calculations into neat compact solutions.
    Here's just a bit i wanted to add.

    For your Q1, i interpreted it as....

    how many different additions of 3 unequal positive non-zero integers sum to 11.

    Hence, i recommended you try

    1+2+(11-3)=11, which is 1+2+8=11

    then 1+3+(11-4)=1+3+7,
    1+4+6=11,
    1+5+5...but this contains 2 fives.

    Starting with 2,
    2+3+6,
    2+4+5,
    2+5+4 is the same as 2+4+5 so move on to 3,
    3+4+4 has two fours, so that's out.

    3+5+3 out
    3+6+2 is ok
    3+7+1 counted

    start with 4
    4+5+2 has been found already as has 4+6+1

    The list then is

    1+2+8
    1+3+7
    1+4+6
    2+3+6
    2+4+5
    3+6+2


    If you meant that

    X_1,\ X_2, X_3 could be variables that can have any value
    from 1 to 9, since 1+1+9=11, so 9 is the largest positive integer useable,

    then we have, starting with 1
    1+1+9
    1+2+8
    1+3+7
    1+4+6
    1+5+5

    Since any variable can be any of these values, these can be arranged in 3! ways, therefore there are 5(3!) = 30 combinations with the digit 1,
    with 119 and 191 and 911 all double-counted, along with 155, 515, 551 double-counted, so we subtract 6.
    30-6 =24.

    starting with 2, not containing 1
    2+2+7
    2+3+6
    2+4+5

    these can also be arranged in (2)3!+3!/2! ways to give 12+3=15 additional combinations.

    starting with 3, not containing 1 or 2

    3+3+5
    3+4+4
    3+5+3 has been counted
    3+6+2 has 2 in it

    so there are 2 new ones which can be arranged in (2)3!/2! ways to account for
    each of the 3 variables having those values.
    this is another 3! combinations =6.

    starting with 4, not containing 1, 2 or 3.

    4+4+3 is out so we have found them all as there are no further combinations due to the fact that we will need smaller digits for the sum.

    Hence the total is 24+15+6=45

    This is the total for this interpretation of the question.

    If the variables must have different values, then we must subtract the three with the repeated digit 1, the three with the repeated digit 2, the three with the repeated digit 3, the three with the repeated digit 4, and the three with the repeated digit 5,
    getting 45-15=30 with all 3 digits different,
    in other words.... combinations of the digits, where the digits are selections of 3 from the 8 available in that case, in such a way that they sum to 11.

    Looking at Soroban's analysis in another way

    *****|*|*****

    The red lines can move 4 places left and 4 places right, giving an additional 8 solutions to the 1 shown....8+1=9 solutions containing the digit 1.

    ****|**|*****

    The red lines can move 3 places left and 4 places right,
    giving 7+1=8 solutions containing the digit 2.

    ****|***|****

    The red lines can move 3 places left and 3 places right giving 6+1=7 solutions with the digit 3.

    Continuing that we get 9+8+7+6+5+4+3+2+1=45 solutions containing repeated digits.
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