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mathmath333

  • one year ago

How many 6-digit numbers contain exactly 4 different digits ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{ How many 6-digit numbers contain exactly 4 different digits}\ ? \hspace{.33em}\\~\\ & a.)\ 4536 \hspace{.33em}\\~\\ & b.)\ 294840 \hspace{.33em}\\~\\ & c.)\ 191520 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. sohailiftikhar
    • one year ago
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    just find the arrangement of 6-digit number and then subtract the arrangement in which they have same 4 digits

  3. mathmath333
    • one year ago
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    9*9*8*7*100-9*9*8*7

  4. sohailiftikhar
    • one year ago
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    yup

  5. mathmath333
    • one year ago
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    but the amswer given is b.)

  6. sohailiftikhar
    • one year ago
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    so you will get b sure

  7. sohailiftikhar
    • one year ago
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    did you find the total arrangement of 6-digit number huh ?

  8. ParthKohli
    • one year ago
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    And yet again, my answer is almost the same. It exceeds the thing only by a little.

  9. ParthKohli
    • one year ago
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    I'm pretty sure I've never ever gotten an exact answer on your questions.

  10. ParthKohli
    • one year ago
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    See, I counted the first case of zero wrongly.

  11. mathmath333
    • one year ago
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    answer is b.)294840

  12. ParthKohli
    • one year ago
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    Here it is redone: If zeroes do not repeat. Case 1.1: Where three digits repeat.\[\binom{5}{1}\times \binom{9}{1}\times \binom{5}{3} \times 8 \times 7 \] Case 1.2: Where two digits repeat.\[\binom{5}{1}\times \binom{9}{2}\times \binom{5}{4}\times \frac{4!}{2!\cdot 2!}\times 7\]

  13. ParthKohli
    • one year ago
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    And I get the exact answer (b) 294840 now.

  14. ParthKohli
    • one year ago
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    First the numbers where zeroes do not occur. Case 1: 3 of one digit, and the rest three places unique. \[\binom{9}{1}\times \binom{6}{3}\times 8 \times 7 \times 6\]Case 2: two of one digit, two of another, the rest two unique.\[\binom{9}{2}\times \binom{6}{4}\times \frac{4!}{2!\cdot 2!} \times 7 \times 6 \] Now the numbers where zeroes *do* occur. Case 1: They do not repeat. Case 1.1: Where three digits repeat.\[\binom{5}{1}\times \binom{9}{1}\times \binom{5}{3} \times 8 \times 7 \]Case 1.2: Where two digits repeat.\[\binom{5}{1}\times \binom{9}{2}\times \binom{5}{4}\times \frac{4!}{2!\cdot 2!}\times 7\] Case 2: They do repeat once.\[\binom{5}{2}\times \binom{9}{1}\times \binom{4}{2}\times 8 \times 7 \]Case 3: They repeat thrice.\[\binom{5}{3}\times9\times8\times 7\]

  15. ParthKohli
    • one year ago
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    I deleted the previous reply. Here's the revised one.

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