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I assumed the way to tackle this problem would be $\frac{n!}{(n-r!)(2!)}$ to account for the two l's

which would result in

$\frac{6!}{3!2!}=60$

However, when I enumerate the different possibilities, I end up with over 60 possibilities- see below:

BAL LSY LBY LAY LAS BLS ASY BAY ABS BYS ALL

ABL LYS YBL LYA LSA SLB SAY BYA SAB BSY LLA

ALB YLS BLY YLA SLA LBS SYA YBA BAS YSB LAL

BLA SLY YLB ALY ALS LSB YSA ABY BSA SYB BLL

LBA SYL BYL AYL SAL BSL YAS YAB ASB SBY LBL

LAB YSL LYB YAL ASL SBL AYS AYB SBA YBS LLB... and so on

How can I calculate this using the permutation formula?

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    $\begingroup$ Not every word has two L's in it. Therefore it doesn't make sense to divide by 2!. $\endgroup$
    – Improve
    Commented Jan 10, 2016 at 3:30
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    $\begingroup$ Count separately ("cases") the words with at most one L and the words with two L's. $\endgroup$ Commented Jan 10, 2016 at 3:31
  • $\begingroup$ OK, so, then I would get $\frac{6!}{2!2!} + \frac{5!}{2!} + 4!$. The first fraction gives me the number of words with two l's, the second the number with one l and the third gives me the number with no l's. The answer is 96. However, I wonder am I overcounting by adding a factorial to account for the case of words with no l's? $\endgroup$
    – A nobody
    Commented Jan 10, 2016 at 3:51
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    $\begingroup$ For no L, I prefer $(4)(3)(2)$, but that happens to be $4!$. For one L, the count is $\binom{4}{2}3!$. For two L, the count is $(4)(3)$. Total $72$. It is somewhat more efficient to count the at most one L (the first two together), which is $(5)(4)(3)$. Then add the $12$ two L's, again getting $72$. $\endgroup$ Commented Jan 10, 2016 at 4:12
  • $\begingroup$ Why is the count for one l $\frac{4!}{2!2!}3!$? We have three possible letters, the first two letters seem to be accounted for by $\frac{4!}{2!2!}$ and the next factorial $3!$ would seem to account for three more letters. Can you explain how you worked out that it was $\frac{4!}{2!2!}3!$? $\endgroup$
    – A nobody
    Commented Jan 10, 2016 at 4:47

3 Answers 3

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Nice question. I would first ignore the double "L" and just compute the ways to permute $3$ out of the $6$ numbers. Using permutations, this is simply $\frac{6!}{3!} = 6 * 5 * 4 = 120.$

Now I account for overcount of the double "L." Notice that I do this after counting the total because those permutations without any L's (like "BAS") are NOT overcounted. We notice that there are $\dbinom{4}{1} * 3 = 12$ permutation forms that contain $2$ L's and $1$ other letter - we first pick a non-L letter, then arrange this letter and two L's into a three-letter permuation (we just have to pick a spot for the non-L letter). We counted all $12$ of these twice. So we need to subtract $12$ from our original count. Now what about $1$ L and two other letters? For this, we notice that there are $\dbinom{4}{2} * 3 * 2 = 36$ permutation forms - we first select two non-L letters from the four available, then arrange these two and the L into a three-letter permutation, which is why we multiply by $3 * 2.$ We have overcounted $12 + 36 = 48$ times.

Our answer is $120 - 48 = \boxed{72}.$ The takeaway: carefully address your overcounts. Using combinations may also be helpful here. Hope this helps!

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    $\begingroup$ Could you elaborate on how you got ${{4}\choose{1}} \times 3$ and ${{4}\choose {2}} \times 3 \times 2$? $\endgroup$
    – Airdish
    Commented Jan 10, 2016 at 5:00
  • $\begingroup$ Sure, let me add some clarifications to my answer. $\endgroup$
    – K. Jiang
    Commented Jan 10, 2016 at 5:03
  • $\begingroup$ Thanks! This is great and all clear except for the way you got $\frac{4!}{2!2!}*3*2$. Can you tell me what each part corresponds to? $\frac{4!}{2!2!}$ corresponds to selecting the two non L letters, right? I would have thought that we only need to divide by 2! once here, leaving 4*3 possible choices for the non L letters. I think the next step would be to multiply by 3 to account for the different placement of L. Is this correct? It leads to the same answer. $\endgroup$
    – A nobody
    Commented Jan 10, 2016 at 5:18
  • $\begingroup$ Glad to help! So when I used $\dbinom{4}{2}$, I am simply selecting two numbers from the four non-L letters, irregardless of the order they're chosen. Now I have two distinct non-L letters and an L to rearrange in order to form a three-letter string. This is why I multiplied by $3 * 2.$ But yes, you're way is perfectly correct, and we do indeed get the exact same answer. $\endgroup$
    – K. Jiang
    Commented Jan 10, 2016 at 5:36
  • $\begingroup$ So when you're looking at the overcounting instances, you only consider one case each right? i.e when you look at two non-L letters, you only consider one of the two L's right? Because the two non-L letter cases for the other L have to be preserved. Otherwise we lose some valid cases... $\endgroup$
    – Airdish
    Commented Jan 10, 2016 at 5:48
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For words that contain at most one L, there are $5!/2! = 60$ ways to permute.

For words that contain two Ls, there are 12 ways to permute

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  • $\begingroup$ How did you get 15 for two l's? $\endgroup$
    – A nobody
    Commented Jan 10, 2016 at 4:41
  • $\begingroup$ And we want three letters you have permuted all $5$ $\endgroup$ Commented Jan 10, 2016 at 5:19
  • $\begingroup$ Ahh....excuse me, there are only 12 ways for two l $\endgroup$ Commented Jan 10, 2016 at 8:45
  • $\begingroup$ For the letter other than the two Ls, you have 4 choices, and for each choice, there are 3 places to place the letter. But anyway Jiang has answered it so well $\endgroup$ Commented Jan 10, 2016 at 8:47
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Number of words with at most 1 L = number of words that can be made from BALSY = 5*4*3 = 60.

Number of words with made with 2 Ls = number of ways to arrange LL and one of BASY = 4*3 = 12. (4 choices for the other letter, and 3 places to put it.)

Number of words with either 0, 1 or 2 Ls = 60 + 12 = 72.

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