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You Have 6 letters X,X,Y,Y,V,Z How many arrangement can be done using 3 letters ?

( like XXV,VZY..ETC)

( I tried N= 6P3/(2!*2!) but I think it's wrong.I was able to do solve it by writing all the cases down but I am looking for a formula to apply )

Thank you for your help

Basil Abu Mallooh
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  • This should be small enough to simply write them all out. A suggestion if you don't feel like doing that: break into cases based on whether or not a letter is repeated. If no letter is repeated then it is as though you just have X,Y,V,Z and you arrange three of those. If a letter is repeated, pick which one. Then pick the location of the non-repeated letter. Then pick which the non-repeated letter is. – JMoravitz Nov 06 '16 at 01:12
  • Thank you for your reply. I was able to do it by writing them all out but I am looking for a formula to apply to make life easier :) – Basil Abu Mallooh Nov 06 '16 at 01:30
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    This answer shows how to use more advanced techniques to determine the arrangements of six letters out of MISSISSIPPI. It is broad enough to be extended to any combination of letters and length of desired arrangement. – Ian Miller Nov 06 '16 at 02:33

2 Answers2

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As I already alluded to in my comment above, you are correct in noticing that the originally proposed method is incorrect.

Often times, for these problem unfortunately, you will need to approach by cases or generating functions. In this specific case we break into two cases:

  • No letter is repeated

  • One letter is repeated

When no letter is repeated, then we are simply working with permutations of size 3 of X,Y,Z,V, which we can see that there are $4^{\underline{3}}=4\cdot 3\cdot 2 = 24$

When one letter is repeated, notice that it will necessarily have two of one letter and one of another letter. Approach via multiplication principle:

  • pick which letter is repeated (x or y) $2$ choices
  • pick which is non-repeated letter (x/y or z or v) $3$ choices
  • pick location of non-repeated letter (first, second or third) $3$ choices

for a total of $2\cdot 3\cdot 3=18$ arrangements of the second type.

There are then a final total of $24+18=42$ total arrangements possible.

JMoravitz
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  • I want to thumbs up your answer but I can't since am new member. Thank you anyways :) – Basil Abu Mallooh Nov 06 '16 at 02:40
  • @BasilBassam ian's comment above is the real winner here as it gives a general method (which was alluded to when I referred to generating functions), though it requires a bit more theory to understand why it works and is more challenging to compute with. – JMoravitz Nov 06 '16 at 02:43
  • I am just having a trouble grasping the concept of why the formula I proposed gives the wrong answer.If we are looking for the number of ways 6 letters arrangement can be made it's 6p6/(2!*2!), but for 3 letters arrangement it doesn't work. I am sorry but I hope you can clarify to me the reason – Basil Abu Mallooh Nov 06 '16 at 02:48
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Re your comment
" I am just having a trouble grasping the concept of why the formula I proposed gives the wrong answer. If we are looking for the number of ways 6 letters arrangement can be made it's 6p6/(2!*2!), but for 3 letters arrangement it doesn't work."

Let us rewrite your formula as $\dfrac{6\cdot5\cdot4}{2!2!}$ to more easily understand the underlying reason.

When you choose only $3$ letters, there can be patterns of various types:

$2-1\;of\;a\; kind,\; e.g. XXY.$ Obviously each pattern of this type will yield $\dfrac{3!}{2!1!}$ results

$1-1-1\;of\;a\;kind\;e.g. XYZ.$ Each pattern of this type will yield $\dfrac{3!}{1!1!1!}$ results.

So you can see that the numerator isn't $^6P_3,$ it is $3!$,
and the denominator isn't a uniform $2!2!$, but varies according to the pattern of the letters chosen.

true blue anil
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