mathmath333
  • mathmath333
How many 6-digit numbers contain exactly 4 different digits ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ How many 6-digit numbers contain exactly 4 different digits}\ ? \hspace{.33em}\\~\\ & a.)\ 4536 \hspace{.33em}\\~\\ & b.)\ 294840 \hspace{.33em}\\~\\ & c.)\ 191520 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)
sohailiftikhar
  • sohailiftikhar
just find the arrangement of 6-digit number and then subtract the arrangement in which they have same 4 digits
mathmath333
  • mathmath333
9*9*8*7*100-9*9*8*7

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sohailiftikhar
  • sohailiftikhar
yup
mathmath333
  • mathmath333
but the amswer given is b.)
sohailiftikhar
  • sohailiftikhar
so you will get b sure
sohailiftikhar
  • sohailiftikhar
did you find the total arrangement of 6-digit number huh ?
ParthKohli
  • ParthKohli
And yet again, my answer is almost the same. It exceeds the thing only by a little.
ParthKohli
  • ParthKohli
I'm pretty sure I've never ever gotten an exact answer on your questions.
ParthKohli
  • ParthKohli
See, I counted the first case of zero wrongly.
mathmath333
  • mathmath333
answer is b.)294840
ParthKohli
  • ParthKohli
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.\[\binom{5}{1}\times \binom{9}{2}\times \binom{5}{4}\times \frac{4!}{2!\cdot 2!}\times 7\]
ParthKohli
  • ParthKohli
And I get the exact answer (b) 294840 now.
ParthKohli
  • ParthKohli
http://www.wolframalpha.com/input/?i=9*20*8*7*6+%2B+%289+choose+2%29*%286+choose+4%29*6*7*6+%2B+5*9*10*56+%2B+5*%289+choose+2%29*5*42+%2B+10*9*6*56+%2B+10*72*7
ParthKohli
  • ParthKohli
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.\[\binom{5}{3}\times9\times8\times 7\]
ParthKohli
  • ParthKohli
I deleted the previous reply. Here's the revised one.

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