# Doubt in reasoning of possible combinations of given digits

How many 3-digit numbers can be made with the digits $0,0,2,4,$ and $6$.

My main concern: is there any particular way to simplify the work without using more advanced counting techniques?

I broke down the counting into three cases:

Case 1: Numbers containing no '0's

The available list is $2,4,$ and $6$. These 3-digit numbers can be arranged in $3!$ ways or 6 ways.

Case 2: Numbers containing at least $1$ zero.

This is more of a side-question, but in this case I reasoned that because there is only two possible locations for zero, counting the amount of combinations for one of these sub cases will allow me to find the total combinations. Since the zero can only go in two locations, swapping with the alternate number (the number in the other possible position for zero) will double the count; thus by just multiplying by $2$, we can determine the amount. I felt that this was a bit "rough" and would probably not work in general. Another approach?

Possible list: $0,2,4,6$ The total number of permutations if we have a zero fixed in one of the two positions is $3 \times2$ because there is $3$ ways of selecting from $2,4,6$ and then $2$ ways from selecting from the remaining $2$.

Then the total combinations is just twice $(3\cdot2)$ or $2\times3\cdot2=12$

Case 3: Two zeros

The two zeros must go in the last two positions, so there are only $3$ numbers of this kind.

The total amount is then $6+12+3=21$

• The way to simplify is dividing the work into parts. For 3 digits : (1) All different (2) 2 similar and 1 different. Again you can subdivide (1) and (2) into parts like: (A) containing one 0 (B)not containing 0 (C) containing 2 zeros – user220382 Nov 30 '16 at 5:11

The simplest way I can see is dividing into $2$ cases ( middle digit zero or non-zero), yielding $3\cdot1\cdot3 + 3\cdot2\cdot2 = 21$