How many 4 digits numbers can be formed from the integer 0-9?

How many 4 digits numbers can be formed from the integer 0-9?

a. if the number must be odd and between 4,200 and 6,800(inclusive) without repetition digits

b. the number must be even and less than 7,200 without repetition digits

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  • 1 decade ago
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    a. 3 x 6 x 8 x 7 - (8 x 7) + 1 = 953

    Because the first digit has 3 possibilities. The second digit would have 7 possibilities (2,3,4,5,6,7 and 8), but one of the possibilities was already used in the first digit... therefore there are exactly 6 possibilities for the second digit. The third digit has 8 possibilities and the fourth digit has 7 possibilities.

    However, so far there are some possibilities I counted that should not be counted. Numbers beginning in 68 such as 6801, 6802, 6803, ..., all the way to 6899, but not including numbers such as 6866 and 6868... because this would be repetition of digits. That is why I subtracted (8x7) because I need to subtract all those possibilities of 68XX... where the third digit has 8 and possibilities and the fourth digit has 7 possibilities. However, now I have eliminated the possibility of 6800, which is to be included. Therefore I added 1.

    b. To do this problem I would draw a tree diagram. Branching out from the beginning I would have the possibilities 7, even #, and odd #. This is just for the selection of the first digit. Also, for this we are going to consider 0 an even number because if the last digit is 0, then that makes the entire number even. Lets call the first branch A, the second B, the third C, and so forth. Lets call the branches that extend from A, AA, AB, AC,...

    A is the first digit being 7 and there is one way that it can happen. B is the first digit being even and there are 4 ways it can happen. C is the first digit being odd and there are 3 ways it can happen.

    AA is the 2nd digit being 1 with one way this can happen

    AB is the 2nd digit being 0 with 1 way.

    (The second digit cannot be 2)

    AAA is an even third digit with 5 ways it can happen.

    From now on I am going to put a (x) where x is the number of ways a conditional event can occur.

    AAB is an odd 3rd digit (3). AAAA is an even 4th digit (4)

    AABA is an even 4th digit (5)

    ABA is an even 3rd digit (4). ABAA is an even 4th digit (3)

    ABB is an odd third digit (4). ABBA is an even 4th digit (4)

    BA is an even 2nd digit (3) BAA is an even 3rd digit (2) BAAA is an even 4th digit (1).

    I hope by now that you get the idea. Every possibility needs to be multiplied out. You multiply going down the tree diagram from left to right. Example (number of ways A can happen) x (number of ways AA can happen) x (number of ways AAA can happen) x (number of ways AAAA can happen)... etc. Then when you have done that for every branch of the tree, add up all the numbers.

    I hope this helps. If you have questions you can put it in the additional information section. I am not going to just give you the answer but I believe the correct answer is a number that seems to count down and whose digits add up to 12. Good luck!

  • Anonymous
    1 decade ago

    Your teacher asked you this in the hope that you would write a program to answer it yourself. Asking here is defeating the object of the exercise - if you don't want to learn, then drop out of school and let someone more grateful take your place.

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