Permutation of 4 Gs out of 5 Gs

Question

Given symbols:-

$GGGGGAAATTTECCS$

No of ways such that exactly 4 Gs out of 5 Gs are together.

I am getting$ \frac{11! × 11}{3!×3!×2!} $.

Some one verify it?

Comments

I am getting $\frac{11!*10}{3!*3!*2!}$

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I am doing

All possible permutation(taking 4 Gs as Single unit and 1G seperately) - Permutation when Single G with 4Gs.

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Doing same.
I am confused in the 2nd part- "Permutation when Single G with 4Gs. "
Taking 4G as a unit. Say this unit as X. Now adding 1 more G to it.
 XG and GX. I am taking this 2 as different permutations.
 ​​​​​​$\frac{12!}{3!*3!*2!} - \frac{2!*11!}{3!*3!*2!}$​

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_A_A_A_4G_T_T_T_E_C_C_S_
At Any given point single G has only 10 out of 12 positions to fill in so
$\frac{11!*10}{3!^{2}*2}$

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But X is also consists of Gs only.

This GGGGG and GGGGG both are same. And thus we need to divide it by 2.

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Jason Yes, You are right. 
_A_A_A G 4G_T_T_T_E_C_C_S_ 
_A_A_A_4G G T_T_T_E_C_C_S_
These two are same. 

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@Tesla! ur method seems working fine but can you point out error in my approarch.

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$11!*10/ 3!^{2} 2!^{2}$

G also 2 so it should also divided by 2 rt

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_A_A_AX4G_T_T_T_E_C_C_S_

_A_A_A_4GXT_T_T_E_C_C_S_

This 2 position are not allowed that's why I excluded it 

At Any given point single G has only 10 out of 12 positions to fill in

 so why divide by 2?

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yes sorry my mistake

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But these 2 positions of G results in same strings.
AAA${\color{Red} G}$GGGGTTTECCS
AAAGGGG${\color{Red} G}$TTTECCS
So we have to divide it by 2.
Please correct me where I am wrong.

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That's why I removed it those 2 string is not allowed only, I already subtracted all those cases

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but only removal will not give correct result

We have to exclude 2 strings and include 1 string

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All 5 together are not allowed, why we would include 1 string? Am I missing something?

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ok, u mean u strongly reject when 5 strings are together

right?

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then why not ans is

$\frac{12!}{3!3!2!}$

that means there will be always atleast a place between GGGG_G

and that place should be fill up with other element

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Yes and that gap can be filled with single or multiple elements

And 12! Can have atleast one permutation where 5g will be present together

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What about treating 4G's as a single unit and considering that as just one G??

it would give 12! / ( 2! * 3! * 3! *2!) permutations.