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

1. 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}}

2. sohailiftikhar

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

3. mathmath333

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

4. sohailiftikhar

yup

5. mathmath333

but the amswer given is b.)

6. sohailiftikhar

so you will get b sure

7. sohailiftikhar

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

8. ParthKohli

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

9. ParthKohli

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

10. ParthKohli

See, I counted the first case of zero wrongly.

11. mathmath333

12. 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.$\binom{5}{1}\times \binom{9}{2}\times \binom{5}{4}\times \frac{4!}{2!\cdot 2!}\times 7$

13. ParthKohli

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

14. ParthKohli
15. 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.$\binom{5}{3}\times9\times8\times 7$

16. ParthKohli

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