# In how many ways can 4-digit numbers be formed using 2,2,8,8 if each digit is used once only?

In how many ways can 4-digit numbers be formed using 2,2,8,8 if each digit is used once only?

I'm confused as to how to solve this problem. If the question was "How many ways can 4-digit numbers be formed using 2 and 8 if each digit can be used any number of times?", the answer should be $$2 \times 2 \times 2 \times 2 = 16$$, since there a 2 possible digits (2 and 8) that can be used for each of the 4 positions.

However I'm not quite sure of how to solve the problem in the title.

• How many ways can you arrange the twos and the eights? Commented Jan 28, 2021 at 14:50

Pascal's triangle is there for a reason. Number completely determined by which two of the 4 slots are assigned the digit "2" : $$~\binom{4}{2}.$$

If the number starts with a $$2$$, there are are three places we can put the other $$2$$. If the number starts with an $$8$$, there are three places we can put the other $$8$$. Since the number must start with a $$2$$ or an $$8$$, there are $$\boxed{3+3=6}$$ possibilities.

The flaw in your reasoning is that your answer doesn't account for the exhaustion of digits, that is, if the first two places are filled with both the $$2$$'s, then there is only one way for the rest, which your solution counts as $$2×2=4$$.

You can rectify it as:

1. If the first two digits are filled with same digit, then total number of ways is $$2×1×1×1=2$$.

2. If the first two digits are distinct, then total number of ways is $$2×2=4$$.

Thus, total number of ways $$=$$ $$\boxed{2+4=6}$$.

ALTERNATIVE SOLUTION:

Think of it in pairs. The two $$2$$'s and $$8$$'s can form four pairs: $$22$$, $$28$$, $$82$$ and $$88$$. All the question remains is to find arrangement of two pairs, which can be done in $${}^4\text{P}_2=12$$ ways. Since this is symmetrical for either $$2$$, the answer is $$\boxed{6}$$.

OR you can think of it, as fixing two $$2$$'s and filling the rest of places with $$8$$'s, giving the same result as above.

Hope this helps. Ask anything if not clear. Have a wonderful day ahead :)

• The notation $~^4P_2$ is for falling factorials and would have equaled $4\times 3 = 12$, not $6$. The related notation similar to yours that you mean to use is the binomial coefficient $~^4C_2$. Personally, I hate both of these notations, preferring $4\frac{2}{~}$ for the falling factorial and $\binom{4}{2}$ for the binomial coefficient. Commented Jan 28, 2021 at 16:02
• @JMoravitz Personally, for universal clarity, I prefer $\frac{4!}{2!}$ over ${}^4P_2$. Commented Jan 28, 2021 at 16:10
• @user2661923 You say that, but being able to write $\dfrac{365\frac{n}{~}}{365^n}$ as the probability in the birthday problem is very aesthetically pleasing. Getting to use the notation $Y\frac{X}{~}$ for the injective functions from $X$ to $Y$ is also quite nice. Commented Jan 28, 2021 at 16:11