Problems from Combinations with repetition

In a bag there are $7$ balls numbered $1$ to $7$ . Without looking, a ball is taken out, the corresponding number is noted, and the ball is returned to the bag. This action is performed three times more (that is to say, in total there are four numbers noted). How many sets of numbers can be obtained by means of this procedure?

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Development:

We have $n = 7$ (since numbered balls from $1$ to $7$ are chosen) and $k = 4$ (because $4$ numbers are written). Also elements can be repeated, because as you leave the ball again inside the bag, you can get this ball back next time.

Also, as it has been said, the order does not matter. So it is a question of a combination with repetition, and therefore according to the expression given previously we have: $C R_{n, k} = \frac{(7 + 4 - 1)!}{(7 - 1)! 4!} = \frac{10!}{6! 4!} = 210$

Solution:

$210$ sets.

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