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Math Help - Help Please Combinations

  1. #1
    Junior Member
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    Exclamation Help Please Combinations

    Hi, I don't know how to solve these problems. Can someone help me?
    1. Of 10 candidates, the first 6 are being interviewed in the morning, and the last 4 in the afternoon. In how many ways can the interviewer arrange a schedule?
    2. How many seven-digit phone numbers can be formed if the first digit cannot be 0 and repetition of digits is not permitted? if repetition of digits is permitted?
    3. Using the letters of the word YOUNG, tell how many different 5-letter combinations are possible if: the first letter must be Y? the vowels and consonants alternate, beginning with a consonant (Y is a consonant here and Y does not have to be first)?
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  2. #2
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    Hello, cuteisa89!

    Of 10 candidates, the first 6 are being interviewed in the morning, and the last 4 in the afternoon.
    In how many ways can the interviewer arrange a schedule?

    I will assume that the order of the candidates is important.

    For the morning interviews, there are P(10,6) = 10򊷸򊐞5 = 151,200 arrangements.

    For the afternoon interviews, there are P(4,4) = 4򉁪1 = 24 arrangements.

    There are: .151,200 x 24 .= .3,628,800 possible schedules.



    How many seven-digit phone numbers can be formed if the first digit cannot be 0
    and (a) repetition of digits is not permitted? . (b) repetition of digits is permitted?

    (a) The first digit can be any of 9 digits (1 to 9).
    The second can be any of the remaining 9 digits.
    The third can be any of the remaining 8 digits.
    The fourth can be any of the remaining 7 digits.
    The fifth can be any of the remaining 6 digits.
    The sixth can be any of the remaining 5 digits.
    The seventh can be any of the remaining 4 digits.

    Answer: .9򊷸򊐞򉩄 .= .544,320


    (b) The first digit can be any of 9 digits (1 to 9)
    The second through the seventh can be any of the 10 digits.

    Answer: .9101010101010 .= .9,000,000



    Using the letters of the word YOUNG, how many different 5-letter combinations
    are possible if: (a) the first letter must be Y?
    (b) the vowels and consonants alternate, beginning with a consonant?
    (Y is a consonant here and Y does not have to be first) What does this mean?

    (a) The first letter is Y.
    Then the letters {O,U,N,G} can be arranged in P(4,4) ways.
    Answer: .4򉁪1 .= .24 combinations.

    (b) The letters must be: consonant-vowel-consonant-vowel-consonant.
    The three consonants can be placed in 3! = 6 ways.
    The two vowels can be be placed in 2! = 2 ways.
    . . There are: .6 x 2 .= .12 combinations.

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