My daughter came house from college yesterday with a ubraintv-jp.com quiz that us couldn\"t work out:

How numerous 9-digit call numbers deserve to we compose from number 1, 2, 3, 4, 5, 5, 5, 5, 5 ?

I make the efforts the simple permutations and combinations rules however am perplexed by the reality that the 5\"s are repeated in the collection (no repetition because that 1 ~ 4). Ns guess the must have actually a reasonably basic answer together it was in sixth grade advanced ubraintv-jp.com quiz.

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Any assist would be evaluate :)


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ns tried the basic permutations and also combinations rules but am perplexed by the truth that the 5\"s are repetitive in the set (no repetition for 1 ~ 4). I guess the must have a reasonably an easy answer together it to be in 6th grade advanced ubraintv-jp.com quiz.

Yes, $9!/5!$ is sensibly simple.

There are $9!$ ways to arrange nine distinct symbols. However, five of the symbols room not distinct. This method you have the right to sort those species into groups of $5!$ each of which room in truth identical. So us divide and calculate: $$\\frac9!5! = 9\\cdot 8\\cdot 7\\cdot 6 \\qquad\\textdistinct arrangements$$

Alternatively: this an outcome can be derived by counting ways to assign locations to the digits 1,2,3,4 and also place the $5$ in the staying positions. $9$ locations for 1 which leaving $8$ places for 2, and also so on. Thus: $9\\cdot 8\\cdot 7\\cdot 6$ distinct arrangements.

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A much easier example to display how this works is counting means to kinds $1,2,3,3$.

There space $4!$, that is $24$, species of these 4 letters; yet they have the right to be sorted into teams of $2!$ each which differ just in the location of the similar threes (although below they room colour coded for convenience*): $$12\\colorred3\\colorblue3, 12\\colorblue3\\colorred3, \\; 1\\colorred32\\colorblue3, 1\\colorblue32\\colorred3, \\; 1\\colorred3\\colorblue32, 1\\colorblue3\\colorred32,\\; \\colorred312\\colorblue3, \\colorblue312\\colorred3,\\; \\colorred31\\colorblue32, \\colorblue31\\colorred32,\\; \\colorred3\\colorblue312, \\colorblue3\\colorred312, \\\\21\\colorred3\\colorblue3, 21\\colorblue3\\colorred3, \\; 2\\colorred31\\colorblue3, 2\\colorblue31\\colorred3, \\; 2\\colorred3\\colorblue31, 2\\colorblue3\\colorred31,\\; \\colorred321\\colorblue3, \\colorblue321\\colorred3,\\; \\colorred32\\colorblue31, \\colorblue32\\colorred31,\\; \\colorred3\\colorblue321, \\colorblue3\\colorred321 $$

unless you are red-blue colour-blind

This question should be an introduction to the subject of multisets and also the multinomial coefficient. In basic when you have actually a arsenal of elements, $a_k$, each with $r_k$ repetitions, it\"s called a multiset. $$\\(a_k, r_k)\\_n = \\(a_1,r_1), (a_2, r_2), ... (a_n, r_n)\\$$

The variety of distinct permutations that this multiset is the multinomial coefficient: $$\\dbinomr_1+r_2+\\cdots+r_nr_1,r_2,\\cdots , r_n=\\frac(r_1+r_2+\\cdots+r_n)!r_1!\\,r_2!\\,\\cdots\\, r_n!$$