The answer is 1560.

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Did the same! Nice!

You must have calculated it like:

$4.\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) .3!\quad +\quad 6.\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) .\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) .2!$

Kartik Sharma
- 6 years, 6 months ago

We regarded as 121 and 112 as same before counting the number of ways because121 can be arranged in 3 ways while 112 can also be arranged in 3 ways. In both the cases, we will get the same formations. Thus, we didn't count it double time!

jaiveer shekhawat
- 6 years, 6 months ago

+1! Nice way :) I did it using complements cuz it was rather confusing for me working out all the repetitions :)

Happy Melodies
- 6 years, 6 months ago

'!' sign is missing in 6!/3!.

A Former Brilliant Member
- 6 years ago

The total number of permutations without the constraint that the permutation should have all digits present is: 4^6

Let us take the complementary set:

Consisting of exactly one digit: 4 Consisting of exactly two digits: (4,2) * (2^6-2) Consisting of exactly three digits: (4,3) * (3^6-3 * (2^6-2) -3)

Hence the answer is:

4^6 - 4 - (4,2) * (2^6-2) - (4,3) * (3^6-3 * (2^6-2) -3) = 1560

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You should explain how you calculate those which consist of exactly N digits.

Note that if you use * as a multiplication sign, leave empty spaces with the next character, otherwise wise we will interpret it as markdown syntax.

Consider the first four numbers. There are only 3 possibilities:

- There are four different numbers
- There are three different numbers
- There are two different numbers

Any less would mean there could not be four numbers total.

Now simply add up the totals for each possibility:

- There are 4! ways for the first four numbers, and 4 ways for each the 5th and 6th, so $24*4*4=384$
- There are 4 possibilities for the number left out, 3 possibilities for the number repeated, and 12 possibilities for the positioning of the two singles among the first four, for a total of $144$ among the first four. Among the last two, there are 7 possibilities (1 if both are the same as we still need to use up one number, and 6 if they aren't, as there are 2 places for the forced number, and 3 choices for the other.) $7*144=1008$
- There are 6 possibilities for the 2 to choose and 14 ways to put them in the first 4 slots ( $2*2*2*2$ minus the 2 ways which only contain one number). There are two permutations of the last 2 digits, for a total of $6*14*2=168$

$384+1008+168=\boxed{1560}$

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There are two cases here. First, there is a digit that is repeated three times and the other three digits are written once in a number. Second, there are two digits which is repeated twice and the other two digits are repeated just once in a number.

First case: Let one of the number is $~~111234$

Each of digit can be repeated maximum three times in a number, so there is $\left(\begin{matrix}4\\1\end{matrix}\right)=4$ ways to select the digit that is repeated three times in a number. Thus, we can form $4\times{\frac{6!}{3!\times{1!}\times{1!}\times{1!}}}=480$ numbers.

Second case: Let one of the umber is $~~334412$

There are two digits can be repeated twice on a number, so there is $\left(\begin{matrix}4\\2\end{matrix}\right)=6$ ways to select the digit that is repeated twice in a number. So, we have $6\times{\frac{6!}{2!\times{2!}\times{1!}\times{1!}}}=1080$ numbers.

Thus, the total number can be formed is $480+1080=1560$ numbers.

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