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

Question

my solution is----> NUMBER can be 777210 this i denote by aaabcd ------ this can be done in ---> 10*1*1*9*8*7*[6!/3!] {1 for a thrice} NUMBER can be 772210 this i denote by aabbcd ------ this can be done in ------------> 10*1*9*1*8*7*[6!/(2!*2!)] {1 for a and b }

and then multiply the answer by 0.9 to remove 0 case at unit place

but correct solution is C(10, 4)* [C(4, 2)* 6!/(2!)2 + C(4, 1)* 6!/3!] 294840

Answer 1

In your aaabcd case, you have a factor of $\frac{6!}{3!}$ which I think is meant to account for "how many ways are there to choose 3 digit positions to be a?"

However, this number should not be $\frac{6!}{3!}$ but instead $\binom 63=\frac{6!}{3!\cdot 3!}$.

If your intention was to make this the multinomial coefficient $\binom{6}{3,1,1,1}$, then that will lead to overcounting, because for example 737527 will be counted both as abacda, acabda, aadabca, and so forth.

Similarly your second term has $\frac{6!}{2!2!}$ which overcounts for the same reason. You want something like $\binom{6}{2,2,1,1}\cdot\frac12\cdot\frac12$ to account for overcounting with the two pairs and the two singletons swapped, respectively.

This gives you the wrong numerical results for an approach that otherwise looks valid.

Answer 2

The trouble is that before permutations in [ ], you are mixing up combinations and permutations.

A double check of the book answer by a slightly different formula is:

[Choose triple/double]$\times$ [Choose the remaining]$\times$ [permute]

which for patterns $AAABCD$ and $AABBCD$ yields

$0.9[\binom{10}1\binom 93\cdot \frac{6!}{3!} + \binom{10}2\binom 82\cdot\frac{6!}{2!2!}] = 294840$