Dare to solve?

Let the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.

So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).

Hence the number of ways to form such 6-digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10 9 8 7 5*13 = 327600 .

But is this the final answer? .....certainly not!!

See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which'd be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.

Thus the answer will be 9 9 8 7 5*13 = 294840

$\dfrac{9}{10}·\begin{pmatrix}10\\4\end{pmatrix}·4!·\begin{Bmatrix}6\\4\end{Bmatrix}=\frac{9}{10}·210·24·65=\boxed{294'840}$

Explanation:

If the 4 digits are given, then, there is $\begin{Bmatrix}6\\4\end{Bmatrix}=65$ ways to display them (see Stirling numbers of the second kind for formula or table of values).

To choose the digits and order them, there's ${10 \choose 4}·4!=210·24=5040$ ways. But we can't have leading 0's, so we either:

• subtract $\frac{1}{10}$ of them (i.e. 504), as that's the probability the number begins with 0 (each digit as the same probability to begin the sequence);
• subtract the ways to start with 0 and pick and order 3 other digits, that is ${9 \choose 3}·3!=84·6=504$ ways.

That makes $(5040-504)·65=\boxed{294'840}$ .

I used a slightly different approach than Parth Lohomi

The number of ways 6 different blank spaces (different cause order matters) to 4 distinct digits is same as the number of onto functions from set off blank spaces to the digits,

hence we have

using inclusion exclusion principle

no. of ways as $10C4\sum { (-1)^{ r } } (4Cr){ (4-r) }^{ 6 }\quad (summation\quad is\quad from\quad r=\quad 0\quad to\quad 4)\quad =\quad 327600$

Now however to remove the ones where first blank space is mapped to 0,

we subtract this case, let the first blank be a, it may be mapped to 0, then the remaining may or may not be mapped to 0 (as its only onto not one one necessarily)

So for the remaining elements we need to find the functions where 0 may or may not be mapped (by someone other than a ) which is simply the sum of the number of onto functions from 5 elements to 4 and from 5 elements to 3 (that is , maybe= yes+no)

so we have

$9C3(\sum { (-1)^{ r } } (4Cr){ (4-r) }^{ 5 }\quad +\quad \sum { (-1)^{ r } } (3Cr){ (3-r) }^{ 5 })\quad =\quad 32760$

subtracting we have 294840,

Thanks for the problem ,this has thoroughly cleared my concept about combinatorics :)

We let the four digits be w,x,y and z. The number can be in the form wwxxyz or wwwxyz. We first find the number of ways to select 4 digits: $10C4=210$ $ways$

Then the number of ways to select the digits to be repeated in each case: $4C2=6$ and $4C1=4$ .

Then the number of ways of arrangements in each case: $\frac{6!}{2!2!}=180$ and $\frac{6!}{3!}=120$

no. of ways = $10C4(4C2)($ $\frac{6!}{2!2!})$ $+$ $10C4(4C1)($ $\frac{6!}{3!})$ = $210(6)(180)$ $+$ $210(4)(120)=226800+100800=327600$

In the above computation, we have used all the $10$ digits including $0$ in selecting the four distinct digits. That means that some of the numbers starts with $0$ . The first digit of the number can be filled with any of the $10$ digits equally. As the digits are treated on an equal footing, we multiply the result by $\frac{9}{10}$ . So

total number of ways = $327600($ $\frac{9}{10})=294840$