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Math Help - Combinatorics.

  1. #1
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    Exclamation Combinatorics.

    How many 4-digit nos. are there which begin with one and have exactly two identical digits? egs. are (1557,1030,1231)
    pls. do give the method applied as well.
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  2. #2
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    Hi Ilsa

    1st case : the two identiical digits are "1"
    Now we have two "1"s and need two more digits.
    a. Can you find number of ways to get these two digits?
    b. Then can you find the number of arrangements that can be made using those 4 digits?

    2nd case : the two identiical digits are "2"
    Now we have two "2"s and need two more digits.
    a. Can you find number of ways to get these two digits?
    b. Then can you find the number of arrangements that can be made using those 4 digits?

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  3. #3
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    Well, it starts with one, so we will first consider when the identical digits are one.
    We have 11AB, 1A1B, and 1AB1. Where, A and B represent the other digits, and I'm also assuming that A is not the same as B (since there are exactly two digits which are identical.) So, A and B can be 2,3,...,9,0. So, A can be 9 possible numbers, while B can be the remaining 8 (since it can't be the same as A).

    So, for 11AB, there are 9*8 = 72 numbers which satisfy your conditions. This will also be the same for 1A1B, and 1AB1. So for the case when the identical digits are 1, there are 72 * 3 = 216 numbers.

    Next, for the case when 1 is not the identical digit, we will have the following three cases: 1AAB, 1ABA, and 1BAA. Again, A can't be the same as B, and neither can be 1. Similar to the previous case, it turns out that there will be 216 of these numbers also.

    So, there will be a total of 216 + 216 = 432 ... so there are 432 such numbers.
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  4. #4
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    Hello Ilsa
    Quote Originally Posted by Ilsa View Post
    How many 4-digit nos. are there which begin with one and have exactly two identical digits? egs. are (1557,1030,1231)
    pls. do give the method applied as well.
    Imagine we have four boxes in a line. The first box on the left contains a 1; so there is no choice involved in putting a number in the first box.

    Now the remaining boxes can either (a) contain another 1 (and this is therefore the repeated digit); or (b) not contain another 1; and therefore it's another digit that is repeated.

    Let's deal with (a) first. We need to do three things:

    1. Decide which box is going to contain the second 1.
    2. Once this has been chosen, decide which digit to put in the first of the two remaining empty boxes.
    3. Decide which digit to put in the remaining empty box.

    So, for step 1, there are 3 choices of box for the second 1.

    For step 2, there are 9 remaining digits to put in the first empty box from the left.

    For step 3, there are 8 digits remaining that can go in the last box.

    So, multiplying these all together, we have 3 x 9 x 8 = 216 ways of completing (a).


    Next, let's consider method (b). Here we need to do three things:

    1. Decide which of the three empty boxes will contain the number that's not going to be repeated.
    2. Decide which digit to put in this box.
    3. Decide which digit - the one that's repeated - is going to go in the remaining two empty boxes.

    Deal with steps 1 - 3 in exactly the same way as we did in (a), multiplying the three separate numbers together to get the total.


    Finally, add your answers to (a) and (b) to get the overall total.


    Grandad
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  5. #5
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    Red face

    thanx!
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