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Suppose we have 25 pegs, split into 5 colours equally i.e 5 red (r),5 green (g),5 blue (b),5 yellow (y) and 5 orange (o). How many ways are there to pick 5 pegs where order doesn't matter i.e choosing r,g,g,b,b is equivalent to choosing g,g,r,b,b?

user341469
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1 Answers1

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Split it into disjoint cases, count the number of options of each case and add up the results:

  • $\color\red0,\color\red 0,\color\red 0,\color\red 0,\color\green 5:\frac{5!}{\color\red4!\color\green1! }= 5$
  • $\color\red0,\color\red 0,\color\red 0,\color\green 1,\color\orange4:\frac{5!}{\color\red3!\color\green1!\color\orange1!}=20$
  • $\color\red0,\color\red 0,\color\red 0,\color\green 2,\color\orange3:\frac{5!}{\color\red3!\color\green1!\color\orange1!}=20$
  • $\color\red0,\color\red 0,\color\green1,\color\green 1,\color\orange3:\frac{5!}{\color\red2!\color\green2!\color\orange1!}=30$
  • $\color\red0,\color\red 0,\color\green1,\color\orange2,\color\orange2:\frac{5!}{\color\red2!\color\green1!\color\orange2!}=30$
  • $\color\red0,\color\green1,\color\green1,\color\green 1,\color\orange2:\frac{5!}{\color\red1!\color\green3!\color\orange1!}=20$
  • $\color\red1,\color\red 1,\color\red 1,\color\red 1,\color\red 1:\frac{5!}{\color\red5! }= 1$

Hence the total number of options is $5+20+20+30+30+20+1=126$.

barak manos
  • 43,109
  • should you multiply the answer by 5, because the last case gives 1 for each colour? – user341469 Nov 20 '16 at 18:13
  • @Think: Since order doesn't matter, there is exactly $1$ case of $5$ different colors. And in any case, you would have to multiply by $5$ only the last case, not the entire answer. – barak manos Nov 20 '16 at 18:21