Find the number of different arrangements of the letters in each word.
MANHATTAN if the word MAN must appear in the arrangement

- anonymous

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- anonymous

7!/(2!*2!)

- anonymous

is that so?no more conditions?

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## More answers

- anonymous

my attempt at this was (1*3*2) * 6! / (3! * 2!)

- anonymous

note:the dividing factorials are for the repitions of individual letters in the arrabgement

- anonymous

where do you get the 7! from?

- anonymous

take "MAN" as one unit (letter ) since we know we can't break that up .i.e it must exist in the final answer then the answer is simply the number permutations.
correction:7!/(3!*2!*2!)

- anonymous

u can arrange 7 letters in 7! ways, the 3 letters of MAN in 3! ways and since arrangements of TT and AA are same so divide by 2!*2! But since MAN should appear as arranged ie A after M and N after A...so we take only one arrangement..and so the answer is 7!/(2!*2!)

- anonymous

no, the relative permutation of "man" cannot be changed.

- anonymous

but there are several possibilities of A and N in the M.A.N. combination
so wouldn't it be 1*3*2 * (6! [the rest of the letters including the extra A's and N's)

- anonymous

that fact leads to the dividing factorials ,but the 3! comes from the fact that there are 3 a's and not that there are 3 alphabets in "man"

- anonymous

druvidfae is wrong..

- anonymous

i just don't get why MAN is assumed to be 1 letter
there's 1 possibly for M and then 3 a's and 2n's
so can't it be 1M 1A 1N
1M 1A 2N
etc.
why isn't it (1*3*2) for the individual possibilities of MAN combinations

- anonymous

see u need MAN to be dere in any arrangement..so it will help if we take it as one letter.

- anonymous

so how would you find the number of different arrangements
with no restrictions
and given the problem: MANHATTAN?

- anonymous

u need MAN to appear as M-A-N! lol :D
I mean u r not allowing ANM or MNA or NAM

- anonymous

9!/(2!3!2!)

- anonymous

i replied tat for no restrictions case..

- anonymous

so then with this restriction it would be... 7!/(3!2!2!)
with 3! accounting for a's, 2! accounting for n's and t's?

- anonymous

no when u r counting with the given restriction u cant permute the letters of MAN..understood?
Use it as another letter which is not there in thee word..like Z..so ur given problem becomes: permute Zhattan

- anonymous

U cant play with the letters of MAN since they need to be arranged..when u account for them as u said then MAN will not appear as MAN in all ur arrangements...

- anonymous

oh okay that makes sense when you change it to a different letter

- anonymous

yah..but u must understand the deep meaning also...why we change it to a different letter!

- anonymous

Cause when account for a's and n's of MAN (which u replied in a post) then u get some arrangements as:
MANHATTAN
NAMHATTAN
HATTAMNAN....
but u want MAN as M-A-N not other arrangements..so u keep that fixed and take Z

- anonymous

because you want the combination M A N to be there in any arrangement so you should assume its one letter to make it a fixed solution?

- anonymous

yups..u got it!

- anonymous

omg thank you so much for being so patient with me :) medal

- anonymous

thnx..ask if u hav any probs..good luck!

- anonymous

ok thx! do you apply the same process when talking about vertices and trying to form triangles and diagonal lines?

- anonymous

no...i dont think so..:) its only for letters
btw...for that kind of problems i know how to do them..ask me if u want!

- anonymous

sorry for imposing :(
the question was:
A regular polygon was 10 vertices. use combinatorics to find:
triangles that could be formed using the vertices
diagonals that could be drawn

- anonymous

a regualr polygon with 10 vertices means no 3 points are in a line..i.e. they are not collinear. So we have triangle from any 3 points from among the 10 points. So that means we have to find the number of ways we can select 3 points from 10 points.i.e. 10C3.This gives the number of triangles.

- anonymous

For diagonals, we man by diagonals any line inside the polygon. So we cant take sides as diagonals. We get a line by joining any 2 points. But evry line is not a diagonal..see that the sides are also lines obtained by joining 2 points...so we find the (total lines)-(total sides) i.e. 10C2-10.which gives the no. of diagonals..hope u understood:)

- anonymous

got it smoochumkisses?

- anonymous

yes it makes sense when you put it conceptually as the points aren't colinear
so then would the diagonal lines question be
10C2 / 2! to account for the diagonal line coming from two directions?

- anonymous

no..it will be 10C2-10.
Not 10C2/2! since in combinations we take one arrangement of a particular pair...u could have said 10P2/2!-10 though..its the same as my answer..

- anonymous

-10? why?

- anonymous

i told u in my post. we must neglect the sides of the polygon. Sides are not diagonals..:)

- anonymous

oh okay, so you don't divide by 2! because it doesn't matter the order just as long as the two points are selected still
but you subtract 10 because the sides can't be diagonals

- anonymous

you got that!! congr8s! :P

- anonymous

you make it easier to understand XD thx u
time to do some binomial expansion problems now hehe
thanks soo much again

- anonymous

ask me if u have any problems for expansions:) happy to help:)

- anonymous

what math are you in at school btw??
this is all pre-calc but im sure you could tell XD

- anonymous

i have passed high school yesterday...:)
got 100/100 in maths..

- anonymous

oh grats on passing :]
omg wicked XD 100

- anonymous

thnx..what math r u in and in which grade?

- anonymous

pre-calc freshman

- anonymous

k.gud! best of luck..

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