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a) How many numbers with seven digits are possible if exactly one digit occurs exactly four times and all other digits are distinct?

b) How many numbers with seven digits are possible if exactly two digits occur exactly two times and all other digits are distinct?

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  • $\begingroup$ What have you tried so far? It's easier to help someone if they describe their attempts or explain what they've done so far rather than just dropping a question off at our doorstep. $\endgroup$
    – Xoque55
    Commented Mar 11, 2014 at 0:01

1 Answer 1

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Hint. It is usually a good idea with counting problems to establish a step-by-step procedure for constructing the situation you want, then find the number of ways to do each step, then combine your answers to get the final result. For your first problem:

(1) Choose the digit which is going to occur four times. . . . . $10$ ways
(2) Choose the four locations in which this digit will occur. . . . . $C(7,4)$ ways
(3) Choose three digits from the remaining six, with repetition disallowed and order important
. . . . . ??? ways.

Final answer, ??? ways. Note that this assumes a number can start with $0$ - if not, you will have to consider various cases.

See if you can do it from here. Good luck!

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  • $\begingroup$ For (1), isn't it 10 ways? Then is (3) 9*8*7? $\endgroup$
    – asdf
    Commented Mar 11, 2014 at 0:21
  • $\begingroup$ @asdf yes of course, didn't read the question carefully. Have fixed it. Thanks. $\endgroup$
    – David
    Commented Mar 11, 2014 at 0:24

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