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Any suggestion on solving this problem to get rid of double/over counting?

The digits $1$, $2$, $2$, $3$, and $4$ are placed on separate cards. How many different $3$-digit numbers can be formed by arranging the cards?

I tried... the case where all the numbers are distinct....like $1,2,3,4,5$ to make $3$-digit # and that is $5 \cdot 4 \cdot 3 = 60$ possibilities. Now I need to correct for over counting as $2$ is a repeated digit. Any help is appreciated.

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  • $\begingroup$ Welcome to MathSE. Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Feb 11 '18 at 18:07
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Consider cases:

Three different numbers are used: There are four ways to select the hundreds digit, three ways to select the tens digit, and two ways to select the units digit. Hence, there are $4 \cdot 3 \cdot 2 = 24$ such numbers.

Two different numbers are used: There must be a repeated $2$. Choose two of the three positions for the two $2$s. Choose one of the other three digits for the free position. There are $\binom{3}{2}\binom{3}{1} = 9$ such numbers.

Total: Since the two cases above are mutually exclusive and exhaustive, there are $24 + 9 = 33$ three-digit numbers that can be formed with the given digits.

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Imagine you have $\color{red}{2}$ and $\color{blue}{2}$, so that 2s are distinct. Every solution with a 2 in it has a corresponding solution where red and blue are swapped. These pairs of solutions should be considered the same; this will get rid of double-counting.

So, to avoid double-counting, you need to add up (1) the number of solutions without any twos, plus (2) half the number of solutions containing twos.

There are $3! = 6$ solutions that don't contain twos, and there are $60-6=54$ solutions containing twos. Hence the correct number of solutions, avoiding double-counting, is:

$$6 + \frac{1}{2}(54) = 6 + 27 = 33$$

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