probability qstn (1 Viewer)

[absta]

In how many ways can the word S I M P L I F I E S be arranged in which the word M I S S will appear?

im not sure wat the right answer is..but this qstn really confuses me

 

[Dreamerish*~]

Wouldn't that just be 7!/2?

It's asking for the number of arrangements if MISS always appears, not for the probability that MISS will appear.

What answer did you get?

EDIT: Actually, 7!/4, since there are two S's in MISS. I'm still not sure.

 

[KFunk]

EDIT: Actually, 7!/4, since there are two S's in MISS. I'm still not sure.

You would only need to divide by 2! for the S's if you first multiplied by 2! to arrange them .

 

[absta]

ok i get that, but would you maybe have to multiply the whole thing by 3 to account for MISS which each of the 3 different I's?

 

[volition]

I'll have a stab at it too:
This isn't a complete answer
I'm guessing you need (4! X 6!)/(2! X 3!)

I think you can treat 'miss' as one group and the other letters as another group (hence the 4! X 6!) and the (2! X 3!) is for the 'repetitions' of S and I (it doesn't matter which one goes where).

This is still incomplete though, because I don't know how to arrange the 'block' of letters (miss) inside the 'big' word.... I think you have to multiply by something to account for that

 

[KFunk]

If the word MISS is to appear then you can't really have MSIS or SMIS so treat [MISS] as a single unit in itself (no arrangements required... if you switch the S's nothing has changed). You're then left with 6 letters, 2 of which are identical. You can treat [MISS] as another individual letter unit which leads you to dreamerish's first result:

7!/2! (the number of permutations of 7 objects with 2 the same)

 

[jake2.0]

i did another way, consider there to be to spaces so _ _ _ _ _ _ _ _ _ _ . lets say that MISS appears at the start so for the first _ there is 1 possible solution (couldn't find right word). for the second _ u choose one of the 3 I's so there are 3 possible solutions. for the third _ u choose one of the 2 S's so u have 2 possible solutions. for the fourth _ u choose the remaining S so u have 1 possible solution. for the rest of the _'s its just 6!

so total ways = 1x3x2x1x6!
= 4320

 

[word.]

i did another way, consider there to be to spaces so _ _ _ _ _ _ _ _ _ _ . lets say that MISS appears at the start so for the first _ there is 1 possible solution (couldn't find right word). for the second _ u choose one of the 3 I's so there are 3 possible solutions. for the third _ u choose one of the 2 S's so u have 2 possible solutions. for the fourth _ u choose the remaining S so u have 1 possible solution. for the rest of the _'s its just 6!

so total ways = 1x3x2x1x6!
= 4320

Um. No.
You didn't take into consideration if MISS wasn't first. And you're supposed to divide from repetition not multiply because they're not distinct. By your logic, arranging the word "MMM" = 3 * 2 * 1 = 6, but really there is only one way because they're not distinct.

KFunk is right. You treat MISS as a single letter.
let [x] = MISS. Number of arrangements of [x]PLIFIE: 7!/2!

For clarification:
In how many ways can the word M I S S I be arranged in which the word M I S S will appear?

Arrangements: 2!: [x]I and I[x], i.e. MISSI and IMISS

 

[speed2]

If the word MISS is to appear then you can't really have MSIS or SMIS so treat [MISS] as a single unit in itself (no arrangements required... if you switch the S's nothing has changed). You're then left with 6 letters, 2 of which are identical. You can treat [MISS] as another individual letter unit which leads you to dreamerish's first result:

7!/2! (the number of permutations of 7 objects with 2 the same)

what 2 are u saying are the same, the 2 S's in MISS or the 2 I's left over?

 

[KFunk]

The I's that are left over.

 

[rnitya_25]

SIMPLIFIES.

3 I's 2 S's, 7 unrepeated letters altogether. if you want miss in every word that you make, then miss can be made in two different ways: miSs and misS, so that 2.

now 2 x 5! i think is rite.

 

[KFunk]

- miss
- miss

^ spot the difference

 

[Dreamerish*~]

- miss
- miss

^ spot the difference

LOL.

isnt it 6!

After you take MISS out, there are 6 letters left over, but when you work out how many combinations there are, you include MISS as one group, so there are 7 "letters".

 

[rnitya_25]

- miss
- miss

^ spot the difference

ok that depends on permutations or combinations

 

[bubu]

isnt it 6!

 

[rama_v]

ok that depends on permutations or combinations

umm... no it doesnt, it applies for both lol
For example, how many ways are there of arranging the letters in the word NINE
its not 4!
its 4!/2!

thats permutations