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我們將相同的藍球3個，\;相同的紅球2個排成一列，\;有幾種方法？

$為了區分，\;我們將藍紅球分別編號，\;這樣就有3個藍球\;\;\;\;\;\;\;\;\;\;\;\;\;\;及2\;個紅球\;\;\;\;\;\;\;\;\;\;\;，\;$

將這些編有號碼的\;5\; 個球作直線排列，\;會有5!\;種排法.

究竟其中有幾種直線排列在抹去球上編號後，\;會和\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;的排列相同呢？

我們列舉如下，\;下列的12種直線排列，\;在抹去球上編號後，\;會形成藍藍藍紅紅的排列.\;

排列數12 是由3!與2!乘積而得的，\;即分別將1 ,2, 3寫在藍球上，\;1, 2寫在紅球上的方法數.

也就是說對每一種排法而言，\;若將三個藍色球與兩個紅色球皆看成不同，\;

則有\;3! ×2! \;種排法.\;事實上，\;這\;12\;種排法在抹去球上編號後都會對應到

同一種排法.\;因此，\;相同的藍球三個與相同的紅球兩個排成一列，\;共有\frac{5!}{3!×2!}種排法.

$Step=%v$

不盡相異物的排列

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A = (-0.7138527581818207, -7.965975159999992)
B: A + (14, 3.6)
C = (-3.7609376781818176, 9.111872879999995)
list1 = Sequence(Circle[A + (0, 1.2 * j), 0.5], j, 1, 6)
list10 = Sequence(Circle[A + (11.8, 1.2 * j), 0.5], j, 1, 6)
list2 = Sequence(Circle[A + (1.2, 1.2 * j), 0.5], j, 1, 6)
list3 = Sequence(Circle[A + (2.4, 1.2 * j), 0.5], j, 1, 6)
list4 = Sequence(Circle[A + (3.6, 1.2 * j), 0.5], j, 1, 6)
list5 = Sequence(Circle[A + (4.8, 1.2 * j), 0.5], j, 1, 6)
list6 = Sequence(Circle[A + (7, 1.2 * j), 0.5], j, 1, 6)
list7 = Sequence(Circle[A + (8.2, 1.2 * j), 0.5], j, 1, 6)
list8 = Sequence(Circle[A + (9.4, 1.2 * j), 0.5], j, 1, 6)
list9 = Sequence(Circle[A + (10.6, 1.2 * j), 0.5], j, 1, 6)
list02 = Sequence(Circle[B + (1.2 * j, 0.6), 0.5], j, 0, 2)
list01 = Sequence(Circle[B + (3.6 + 1.2 * j, 0.6), 0.5], j, 0, 1)
s = 0 (slider from 0 to 5, step 1)
lista = Sequence(A + (0, 1.2 * j), j, 1, 6)
listb = Sequence(A + (1.2, 1.2 * j), j, 1, 6)
listc = Sequence(A + (2.4, 1.2 * j), j, 1, 6)
listd = Sequence(A + (3.6, 1.2 * j), j, 1, 6)
liste = Sequence(A + (4.8, 1.2 * j), j, 1, 6)
listf = Sequence(A + (7, 1.2 * j), j, 1, 6)
listg = Sequence(A + (8.2, 1.2 * j), j, 1, 6)
listh = Sequence(A + (9.4, 1.2 * j), j, 1, 6)
listi = Sequence(A + (10.6, 1.2 * j), j, 1, 6)
listj = Sequence(A + (11.8, 1.2 * j), j, 1, 6)
D: C + (16.3, -0.5)
c = Circle(D, 0.3)
D_2: D + (0.8, 0)
d = Circle(D_2, 0.3)
D_3: D_2 + (0.8, 0)
e = Circle(D_3, 0.3)
D_4: D_3 + (4.4, 0)
D_5: D_4 + (0.8, 0)
f = Circle(D_4, 0.3)
g = Circle(D_5, 0.3)
D_6: D + (-0.85, -1.9)
D_7: D_6 + (0.8, 0)
D_8: D_7 + (0.8, 0)
D_9: D_8 + (0.8, 0)
D_a: D_9 + (0.8, 0)
h = Circle(D_6, 0.3)
p = Circle(D_7, 0.3)
q = Circle(D_8, 0.3)
r = Circle(D_9, 0.3)
t = Circle(D_a, 0.3)