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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.

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  • $\begingroup$ How many ways can you arrange the twos and the eights? $\endgroup$
    – Joshua Wang
    Jan 28, 2021 at 14:50

3 Answers 3

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

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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.

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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 :)

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  • $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Jan 28, 2021 at 16:02
  • $\begingroup$ @JMoravitz Personally, for universal clarity, I prefer $\frac{4!}{2!}$ over ${}^4P_2$. $\endgroup$
    – user2661923
    Jan 28, 2021 at 16:10
  • $\begingroup$ @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. $\endgroup$
    – JMoravitz
    Jan 28, 2021 at 16:11

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