Question

In how many ways can the letters in the word BUTTERFLY be arranged if the letters are taken 5 at a time

Answer 1

9 letters
T is twice

Answer 2

It'd be 9p5 ----> 9 Permutation 5

\(\Large \color{MidnightBlue}{\Rightarrow {9! \over 4!} }\)

Answer 3

i thought it would be (9!/2!)/(9-5)!

Answer 4

then divide that by 2!

Answer 5

\(\Large \color{MidnightBlue}{\Rightarrow {9 * 8*7*6*5*\cancel{4!} \over \cancel{4!}} }\)

Answer 6

The question doesn't mention that @PaxPolaris

Answer 7

becausr there are 9 letter and u only choe 5

Answer 8

\(\Large \color{MidnightBlue}{\Rightarrow 72 * 42 * 5 }\)

Answer 9

@Curry no, in this case, the ORDER matters.

Answer 10

wait ... both are wrong

the Order of the 2 T's doesn't matter only if T is chosen

Answer 11

I meant that if you have to find the ways of arranging, you use ONLY permutations.

Answer 12

ye but the reason i divided the 9! by 2! intially is because we have 2 T

Answer 13

so am i wrong ao wat is the right anwer and why?

Answer 14

I'm getting it as 72 * 210

Answer 15

You can't just choose combinations in this case.

Answer 16

If both T are chosen:\[\Large _7P_3 \times( _5C_2)\]
If 1 T is chosen: \[\Large _7P_4 \times( _5C_1)\]
If no T are chosen: \[\Large _7P_5\]

add them all up for your answer

Answer 17

@PaxPolaris Hey, we're talking about all the letters. We don't have to worry about the T's.

Answer 18

but it is however many different combo u can make

Answer 19

Yes @Curry , so, you will use PERMUTATIONS.

Answer 20

@ParthKohli yes we do:
let's look at a simpler problem: how many different way to arrange: ATA

ATA
TAA
AAT

that's 3 ... and not 3!

Answer 21

this is not simply permutation ... like arranging ABC

Answer 22

3p3 \(\Large \color{MidnightBlue}{\Rightarrow {3! \over (3 - 3)! } }\)

Answer 23

Oh yes.

Answer 24

But, @PaxPolaris this problem is more about the permutations.

Answer 25

so wwwhat i the right anwer

Answer 26

72 * 210, as I said.

Answer 27

yes but because there are repeating letterd you have to use both ... permutaions and combinations

Answer 28

no thats too high ...

Answer 29

I dealt with having 2T, 1T and 0T seperately and added them up ...

\[\Large (_7P_3 \times_5C_2) + 5(_7P_4)+(_7P_5)\]

Answer 30

that gives:
\[\huge8820\]

Answer 31

PaxPolaris is correct

Answer 32

Oh oh @Zarkon sorry