Permutations and Combinations in arranging letters

Question

Number of ways to arrange the letters $(A,A,A,A,A,B,B,B,C,C,C,D,E,E,F)$ such that no two $C's$ are together ?

My solution:

I calculated all possibilities with CC together i.e $\dfrac{14!}{5!3!2!}$ and subtracted it from total i.e $\dfrac{15!}{5!3!3!2!}$.

But my answer is wrong. Why?

There are three types where 2 or more C are together, namely CCbbbbC,CbbbbCC,CCC. (the bbbb meaning at least one non-C between. — coffeemath, Dec 16 '16 at 14:50
@coffeemath The * is used for italics. Use \* instead, or surround the whole expression with backticks ( ` ) to get, for instance, CC*C — Arthur, Dec 16 '16 at 14:51
@Arthur Thanks-- didn't know that. (too late to change comment..) — coffeemath, Dec 16 '16 at 14:54

Arthur · Answer 1 · 2016-12-16T14:56:33.260

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Because the combinations where all three C are adjacent were mishandled: each case of CC C was counted separately from all the cases of C CC, when in reality they're the same. You therefore need to subtract once all different combinations that have all three C's together to get the correct number of combinations where at least two C's are together.

edited Dec 16 '16 at 14:56

answered Dec 16 '16 at 14:45

Arthur

199,419

score 0 · Answer 2 · answered Dec 16 '16 at 14:50

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HINT: Let us denote the three $C$'s by $C_1, C_2, C_3$. There are $4$ possible ways of these $C$'s coming together: $C_1C_2, C_2C_3, C_1C_3, C_1C_2C_3$. I hope you will now be able to continue.

answered Dec 16 '16 at 14:50

Kanwaljit Singh · Answer 3 · 2016-12-17T05:05:05.153

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12 alphabets without C's can be arranged in 12! ways. ...(A)

We can arrange 3 C's in any of the 13 (=12+1) positions marked as * below.

$* 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * $

(where 1, 2… 13 represent remaining alphabets)

This can be done in C(13,3) ways. ......(B)

From (A) and (B), required number of ways

$= \frac{12! × C(13,3)}{5! × 3! ×2!}$

$= \frac{12! × 13!}{10! × 3! × 5! × 3! ×2!}$

edited Dec 17 '16 at 05:05

answered Dec 16 '16 at 15:25

Kanwaljit Singh

8,610

If you are using the "gap" method, you would first permute the non-C's, and place the C's in the gaps. Furthermore, you appear to have taken the C's as distinct instead of identical. – true blue anil Dec 17 '16 at 04:21
@true blue anil so what changes I have to made? – Kanwaljit Singh Dec 17 '16 at 04:25
$\dfrac{12!}{5!3!1!2!1!}\times\binom{13}{3}$ – true blue anil Dec 17 '16 at 04:30
Why 12! ? We have 13 places to fill C's. – Kanwaljit Singh Dec 17 '16 at 04:47
The first part is permuting the non-C's. The second part is placing the 3 C's in any 3 of 13 places available – true blue anil Dec 17 '16 at 04:49
Ok got it. Thank u so much. – Kanwaljit Singh Dec 17 '16 at 04:51
Is it fine now? – Kanwaljit Singh Dec 17 '16 at 04:52
The C's are identical, so $C(13,3) \equiv \binom{13}3$, not $P(13,3)$ – true blue anil Dec 17 '16 at 05:00
Thank you. And if still any problem in my answer you can edit my answer or post your own answer. After all we all have same motive to help someone. – Kanwaljit Singh Dec 17 '16 at 05:06

Permutations and Combinations in arranging letters

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