Question

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

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

7!/(2!*2!)

Answer 3

is that so?no more conditions?

Answer 4

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

Answer 5

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

Answer 6

where do you get the 7! from?

Answer 7

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!)

Answer 8

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!)

Answer 9

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

Answer 10

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)

Answer 11

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"

Answer 12

druvidfae is wrong..

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

9!/(2!3!2!)

Answer 18

i replied tat for no restrictions case..

Answer 19

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?

Answer 20

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

Answer 21

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...

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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?

Answer 26

yups..u got it!

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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!

Answer 31

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

Answer 32

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.

Answer 33

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:)

Answer 34

got it smoochumkisses?

Answer 35

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?

Answer 36

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..

Answer 37

-10? why?

Answer 38

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

Answer 39

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

Answer 40

you got that!! congr8s! :P

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

oh grats on passing :]
omg wicked XD 100

Answer 46

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

Answer 47

pre-calc freshman

Answer 48

k.gud! best of luck..