Self doubt

Question

There are 10 different balls in such way that 6 balls are white and 4 balls are black. How many different arrangements are possible such way that black ball placed before the white ball ? 

Answer 1

We have $10$ different balls.

Out of them, $6$ is White & $4$ are Black.

We have to place the Black balls before White balls.

We''ll first arrange Black balls.

Now, the Black balls can itself arrange in $4!$ ways.

Now, the second task is to place the White balls after Black balls.

We can see in above image we can't use the first place to fulfill the given criteria.

So, we have $6$ places, & we can place the White balls into these $6$ places.

Now, the $6$ white balls can itself arrange in $6!$ ways.

So, the answer will be $6! \times 4!$ ways = $17280$ ways.

Comments

Suppose the black balls are B1,B2,B3,B4

and the white balls are W1,W2,W3,W4,W5,W6

I am not getting your question if your question is that the black ball suppose B2 is always placed before the white ball W1 then your answer is (10!/2)

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But the question is to place the black balls before white balls, there are no mentioning which black ball is placed before which white ball

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The question is that

How many different arrangements are possible such $that BLACK BALL$is placed before $the WHITE BALL$

there is no mention of black balls or white balls

May be u r right but my doubt is justified