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How many permutations of $A_1, A_2, A_3, A_4, B_1, B_2, B_3$ have the $A$'s in ascending order and $B$'s in ascending order. (i.e, $A_1, A_2, B_1, A_3, B_2, B_3, A_4 $). Can the solution be extended?($A_1,\dots,A_n,B_1,\dots,B_m,C_1,\dots, C_s$)

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2 Answers 2

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See all the elements as if all were dots $\bullet$. Then , you select $3$ of them, and they will be the $A$'s of your arrangement. Since your $A$'s must be in ascending order, you must put them in order. The rest dots willbe the $B$'s in ascending order. Example:

$$ \bullet\bullet\bullet\bullet\bullet\bullet\bullet$$

I select the ones with stars

$$ \bullet\star\star\bullet\star\star\bullet$$

Then the arrangement is

$$B_1,A_1,A_2,B_2,A_3,A_4,B_3$$

The number of arrangements will be

$$\binom{7}{4}=\binom{7}{3}=\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}= 7\cdot 5= 35$$

For the case of $A_1,\dots,A_n,B_1,\dots B_s$, you can do the same: Consider all elements as points, and then select $n$ elements, that will be your $A$'s. Then order them is ascending order, and put the $A$'s. The rest of elements are $B$'s so put them in ascending order too and you will have all the arrangements in this way. The number of arrangements will be

$$\binom{n+s}{n}=\binom{n+s}{s}$$

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  • $\begingroup$ @ MonsieurGalois, Can the solution be extended?($A_1,\dots,A_n,B_1,\dots,B_m,C_1,\dots, C_s$) $\endgroup$
    – Angel
    Nov 19, 2016 at 18:40
  • $\begingroup$ @Angel Of course, you can do it like $\binom{n+s}{n}$. Is the same method. $\endgroup$
    – iam_agf
    Nov 19, 2016 at 18:41
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Order the As first: $A_1,A_2,A_3,A_4$.

Now, $B_2$ can be placed at any one of $5$ slots.

When $B_2$ is placed at slot #$n$:

  • $B_1$ can be placed at any one of $n$ slots
  • $B_3$ can be placed at any one of $6-n$ slots

Hence the total number of ways to order the Bs is $\sum\limits_{n=1}^{5}n(6-n)=35$.

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  • $\begingroup$ @ barak, Can the solution be extended?($A_1,\dots,A_n,B_1,\dots,B_m,C_1,\dots, C_s$) $\endgroup$
    – Angel
    Nov 19, 2016 at 18:39

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