5 of the 9 letters of the word MINCEMEAT are selected.

Find the number of possible selections which contain exactly 1 M and 1 E.

Find the number of possible selections which contain at least 1 M and at least 1 E.

tried many times ples help thx lamo

Guest Sep 22, 2018

#1**+1 **

5 of the 9 letters of the word MINCEMEAT are selected.

MM EE INCAT

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Find the number of possible selections which contain exactly 1 M and 1 E.

Does order count, you do not say? I'll assume not. I will also assume that both Ms and both Es are identical so that choosing one is exacly the same as choosing the other. So I am counting the number of distinct outcomes. This may not be what the writer of the question intended.

So i must chose 3 letters from INCAT 5C2 = 10 (edited after the interaction below) distinct combinations meet this description.

Of course if you were doing it for a probability question then some of these combinations are more likely to occur than others.

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Find the number of possible selections which contain at least 1 M and at least 1 E.

M and E and 3 others not M or E 5C3

M and E and M and 2 others not M or E 5C2

M and E and E and 2 others not M or E 5C2

M and E and E and another M and 1 others not M or E 5C1

Add those up to get the number of distict combinations.

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Melody
Sep 22, 2018

#6**+1 **

Melody: I think what he means is the answer to the first part is 10. And it is arrived at by 5C3 = 10

And the answer to the second part is 35. And it is arrived by [5C1 + 5C2 +5C2 + 5C3]=5 + 10 + 10 + 10 =35.

Guest Sep 23, 2018

#7**+1 **

Thanks Guest, now the first one makes total sense.

There are 3 to chose from out of 5 then add in the M and the E.

So, yea that is the correct number of possible arrangements, I'm in total agreement.

For the second question it is saying you can have

ME then chose 3 out of the 5 5C3 or

MEE then chose 2 out of the 5 5C2 or

MME then chose 2 out of the 5 5C2 or

MMEE then chose 1 out of the 5 5C1

Add those all up and you have the number of ways.

Yes that makes total sense too. I just realized that this second answer is exactly what I did in the first place LOL.

Melody
Sep 23, 2018

#8**+1 **

Hi Melody, how do you justify this statement?

*“Of course if you were doing it for a probability question then some of these combinations are more likely to occur than others.”*

Why would a question seeking a probability solution change the enumeration, that is make some combinations more likely to occur?

GA

GingerAle
Sep 23, 2018

#9**+2 **

Hi Ginger,

5 of the 9 letters of the word MINCEMEAT are selected.

Find the number of possible selections which contain exactly 1 M and 1 E.

I just mean that if 5 letters are selected at random then the prob of drawing MECAT is higher then the prob of drawing INCAT (becasue there are 2Ms and 2Es to chose from)

So saying there are 10 distinct outcomes is correct but each one does not have the same probability of occuring as other possible outcomes.

eg

What is the probability of chosing a selection of 5 letters that contain exactly one M and one E

We have determined that there are 10 such unique combinations

How many combinations (not unique) are there altogether. (think in terms of M1, M2, E1, E2)

9C5 = 126

Now how many of these have exactly one M and one E

M1 OR M2, E1 or E2, plus 3 more.

2*2*5C3 = 40

So I think that if 5 letters are selected at random then the prob of having exactly one M and one E is 40/126 = 20/63

Melody
Sep 23, 2018

#10**+1 **

Thank you! That makes a big banana-boat load of sense!

The probability of their occurrence is higher than the final observable count.

While this seems to the case for elements selected from a single (dependent) set, when elements are selected from independent sets, such as “pick 3” lottery based questions, the probability directly reflects the enumeration, when a player “boxes” (permutes) a selection where (2) of the numbers are the same.

Dissection of the statistical processes clearly shows why duplicate numbers require **division **to calculate the probability of success.

Your analysis of a dependent set shows why duplicate numbers require **multiplication **to calculate the probability of success.

GA

GingerAle
Sep 24, 2018