Question

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

Answer 1

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

Answer 2

just find the arrangement of 6-digit number and then subtract the arrangement in which they have same 4 digits

Answer 3

9*9*8*7*100-9*9*8*7

Answer 4

yup

Answer 5

but the amswer given is b.)

Answer 6

so you will get b sure

Answer 7

did you find the total arrangement of 6-digit number huh ?

Answer 8

And yet again, my answer is almost the same. It exceeds the thing only by a little.

Answer 9

I'm pretty sure I've never ever gotten an exact answer on your questions.

Answer 10

See, I counted the first case of zero wrongly.

Answer 11

answer is b.)294840

Answer 12

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.\[\binom{5}{1}\times \binom{9}{2}\times \binom{5}{4}\times \frac{4!}{2!\cdot 2!}\times 7\]

Answer 13

And I get the exact answer (b) 294840 now.

Answer 14

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

Answer 15

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.\[\binom{5}{3}\times9\times8\times 7\]

Answer 16

I deleted the previous reply. Here's the revised one.