$4$ digit combinations from $1,1,2,3,4,5$

Question

How many different $4$ digit numbers can be made from the numbers $1,1,2,3,4,5$? Note that there are two ones. I have looked at other examples but find them difficult to follow.

Answer 1

Split it into disjoint cases:

• The number of combinations with $\color\red0$ ones, which is $\binom{4}{4-\color\red0}\cdot\frac{4!}{\color\red0!}=24$
• The number of combinations with $\color\red1$ ones, which is $\binom{4}{4-\color\red1}\cdot\frac{4!}{\color\red1!}=96$
• The number of combinations with $\color\red2$ ones, which is $\binom{4}{4-\color\red2}\cdot\frac{4!}{\color\red2!}=72$

Answer 2

Put the extra $1$ to one side for a moment. Then you have $5\times 4 \times 3 \times 2 = 120$ arrangements of $4$ digits from the set.

Now consider choosing both the $1$s. Then you have $4\times 3$ ordered choices for the other two digits, and for each choice you have $6$ place options to combine them with $1$s, giving $4\times 3 \times 6 = 72$ orderings.

Thus the total is $120+72=192$ options in total.

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That "$6$ place options" comes from $\binom 42$, read as "$4$ choose $2$", and calculated as $\frac{4!}{(4-2)!2!}$. It is a binomial coefficient, very useful for this kind of combinatorics question.

Answer 3

You may separate in 3 cases:

1st case:1 don't appear:you have only 2,3,4,5 so you have only 4! numbers: 2nd case:exist exactly one 1: you have 4 numbers to choose 3, so it's ${4 \choose 3} \cdot 4!$ 3rd case:exist exactly two 1:${4 \choose 2}\cdot \frac{4!}{2!}$

So the total of numbers is the sum of the cases

here ${a \choose b} =\frac{a!}{(b!)\cdot(a-b)!}$