I agree.
]]>You can't do that as that allows for possibilities in which there's a girl at both ends
Not at the same time!
Exactly. Jacks's method allows girls to be at boths ends at the same tiem that was where his proof went wrong.
He wanted us to tell him where his proof went wrong, didn't he?
]]>You can't do that as that allows for possibilities in which there's a girl at both ends
Not at the same time!
]]>but i did not understand the meaning of
You can't do that as that allows for possibilities in which there's a girl at both ends
]]>But where i have done mistake in my solution.
Your mistake is in
now arrange girls in 7 gaps ,
X_X_X_X_X_X_
or
_X_X_X_X_X_X
Each set has
ways; hence there are ways altogether.]]>_ G _ G _ G _ G _ G _ G _ = 6!
In the first _ 6 boys can go
_ G (6 boys ) G _ G _ G _ G _ G _
In the second 5 boys
_ G (6 boys ) G (5 boys ) G _ G _ G _ G _
all the way down to
_ G (6 boys ) G (5 boys ) G (4 boys) G (3 boys) G (2 boys) G _
That can be done in 6 x 5 x 4 x 3 x 2 = 6!
So far we have 6! * 6!, now the last two slots
_ G (6 boys ) G (5 boys ) G (4 boys) G (3 boys) G (2 boys) G _
the one remaining boy can go in 2 ways.
2 * 6! * 6!
]]>But where i have done mistake in my solution.
]]>I am getting 2 * 6! * 6! as the answer. Provided we are assuming the girls and boys are distinct which is reasonable.
]]>my solution::
Using gap method::
here X = denote boys and _ = denote girl
Then X _ X _ X _ X _ X _ X _
first we can arrange boys , which can be done in 6! ways
now arrange girls in 7 gaps ,
for that first we select 6 place out of 7 which can be done in C (7,6) ways and now arrange these 6 girls
Which can we done in C(7,6) * 6! = 7!
So total no. of ways is = 6! * 7!
but answer is = 2*6! *6!
so where i have done mistake
Thanks
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