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The problem is as follows:

The number of ways in which six $A$'s and four $B$'s can be arranged, so that no two $B$'s are together ?

So, we have $AAAAAABBBB$ here.

I tried as follows:

Fixing $A$'s first : $A-A-A-A-A-$

The remaining letters are $ABBBB$

Now arranging these five letters in those five $-$'s : $\frac{5!}{4!}=5$

But here the $A$'s never take even places, so one more criteria is there: $-A-A-A-A-A$

So another $\frac{5!}{4!}=5$ ways for this.

Thus, the total number of ways must be $5+5=10$

But the answer says $35$ !

Can anyone explain ?

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You have $$-A-A-A-A-A-A-$$ and the $B$'s can go in the spaces shown. $B_1$ has 7 choices, $B_2$ has $6$, then $5$ and then $4$. This gives $7\times6\times5\times4$ ways to arrange these.

Then notice that $B_1$ is the same as any other $B$, and so to account for this, you divide by $4!$ since there are $4$ indistinguishable letters. So the final answer is $$\frac{7\times6\times5\times4}{4!}=\frac{7\times6\times5\times4}{24}=7\times 5=35$$

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  • $\begingroup$ Why not just choose $4$ of the $7$ spaces indicated by dashes in your diagram for the $B$s? $\endgroup$ Nov 18 '17 at 12:51
  • $\begingroup$ @N.F.Taussig It's the same thing, but I feel like seeing it like this makes it easier for people who are confused by it to understand. $\endgroup$
    – John Doe
    Nov 18 '17 at 12:53
  • $\begingroup$ So the point is, we don''t have fixed number of slots like I assumed (10) ? Right ? $\endgroup$
    – User9523
    Nov 18 '17 at 12:53
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    $\begingroup$ @User9523 yes, this method (often called stars and bars method) works based on having some things fixed (here the $A$'s) and seeing where the other things can go (the $B$'s). $\endgroup$
    – John Doe
    Nov 18 '17 at 12:55
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Your approach is very circuitous. Does not go well. Best way to think is you have to insert B's as a single into slots surrounded by As.

$$\square A \square A \square A \square A \square A \square A \square$$

We want to fill $4$ B around $6$ As, then we have a total number of arrangements that is ${(6+1)\choose 4}$. This way there are not two B's together.

Number of ways = ${7\choose4} = \frac{7!}{3!4!} = 35$

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  • $\begingroup$ What about the logic above ? Is it wrong ? 2 $B$'s won't come together there also ? (I guess) $\endgroup$
    – User9523
    Nov 18 '17 at 12:49
  • $\begingroup$ An illustration would be helpful, so that the OP understands that you are selecting four of the seven spaces created by placing the six As in a row. $\endgroup$ Nov 18 '17 at 12:49
  • $\begingroup$ @N F Taussig, I am getting out of my house now . Would you want to edit my answer with illustration $\endgroup$ Nov 18 '17 at 12:51
  • $\begingroup$ @SatishRamanathan I added the diagram. $\endgroup$ Nov 18 '17 at 12:54
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    $\begingroup$ @User9523 what was wrong with your logic is the equally spaced $A$'s. The fact that you had included 1 of the $A$'s with the 4 $B$'s to place in the gaps meant that you would always have a line of 3 $A$'s in a row. Your method did not count the permutations which do not have this.$$B\color{red}AB\color{red}AB\color{red}AA\color{red}BA\color{red}A$$doesn't fit the first form you had of $-A-A-A-A-A$ due to there being a $\color{red}B$.$$\color{red}BA\color{red}BA\color{red}BA\color{red}AB\color{red}AA$$doesn't fit the second form you had of $A-A-A-A-A-$ due to there being 3 $\color{red}B$'s. $\endgroup$
    – John Doe
    Nov 18 '17 at 13:07

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