Question

We have 12 letters-
4 A's
3 B's
2 C's and
D,E&F
The are to be arranged in a row such that the C's are separated from one another.
The total number of ways this can be done is-

Answer 1

the rest of the letters might as well be all the same ....

Answer 2

12 spots, 2 cs, and need to be separte

Answer 3

cxcxxxxxxxxx
cxxcxxxxxxxx
.... there are 9 of those :)

move over by 1

xcxcxxxxxxx

tthere are 8 of those

9+8+7+6 .... is what im thinking

Answer 4

err, starts with 10 ways eh

Answer 5

_A_A_A_A_B_B_B_D_E_F_


the As Bs and D,E,F can be arranged in
\[\frac{10!}{4!*3!}\] ways...(1)

now we can fill 12 remaining spots with 3Cs.....is 12C3 ways.....(2)

multiply (1)(2)

Answer 6

sorry (2) its in 12C2 ways...

Answer 7

my idea takes me too:

10(10 + 1)/2 = 5(11) = 55 ways ....

Answer 8

answer in the book is 1386000 ... I'm not quite getting there

Answer 9

im of course ignoring that some letters are positioned in other manners, so it prolly a bit low lol

Answer 10

number of different ways to arrange the nonC letter ...

times 55 ways to put in the cs afterwards ... maybe

Answer 11

you can also try solving like this :

1) find total number of ways of arranging the given 12 letters
2) find total number of ways of arranging 12 letters such that 2Cs are together
3) subtract 2 from 1

Answer 12

1) 12! / 4!3!2!
2) 11! / 4!3!
3) ?

Answer 13

\[\frac{n}{a!b!c!...}\]

Answer 14

\[\frac{10!}{4!3!2!}*55\]

lets see if the wolf likes it lol

Answer 15

693 000 aint what the book says, so ugh!!

Answer 16

@amistre64 logic is correct...

its

\(\frac{10!}{4!3!}*55\)


2! shouldnt be there in the bottom as we fixed c's to start with !!

Answer 17

@amistre64 u should remove the 2!....in the denominator...