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 :)

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-blindThis 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: $$\dbinomr_1+r_2+\cdots+r_nr_1,r_2,\cdots , r_n=\frac(r_1+r_2+\cdots+r_n)!r_1!\,r_2!\,\cdots\, r_n!$$