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mathmath333
 one year ago
How many 6digit numbers contain exactly 4 different digits ?
mathmath333
 one year ago
How many 6digit numbers contain exactly 4 different digits ?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{ How many 6digit numbers contain exactly 4 different digits}\ ? \hspace{.33em}\\~\\ & a.)\ 4536 \hspace{.33em}\\~\\ & b.)\ 294840 \hspace{.33em}\\~\\ & c.)\ 191520 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just find the arrangement of 6digit number and then subtract the arrangement in which they have same 4 digits

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.09*9*8*7*1009*9*8*7

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but the amswer given is b.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you will get b sure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did you find the total arrangement of 6digit number huh ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1And yet again, my answer is almost the same. It exceeds the thing only by a little.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I'm pretty sure I've never ever gotten an exact answer on your questions.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1See, I counted the first case of zero wrongly.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0answer is b.)294840

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Here 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
 one year ago
Best ResponseYou've already chosen the best response.1And I get the exact answer (b) 294840 now.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1First 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
 one year ago
Best ResponseYou've already chosen the best response.1I deleted the previous reply. Here's the revised one.
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