Hi, i'm a bit confused over the steps used to solve the following question.
How many bit strings of length 8 are there where the `1' bits are not next to each other?
Shouldn't the answer be 4! * 9C5 * 5! ? I have this understanding based on the questions below.
How many ways are there for 10 women and 6 men to sit in a row so that no two men
are next to each other?
Ans : 10! ways to arrange the woman * (11C6) * 6! to insert the man.
How many bit strings of length 8 are there where the `1' bits are not next to each other?
Shouldn't the answer be 4! * 9C5 * 5! ? I have this understanding based on the questions below.
How many ways are there for 10 women and 6 men to sit in a row so that no two men
are next to each other?
Ans : 10! ways to arrange the woman * (11C6) * 6! to insert the man.
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one case explained should suffice
let there be six 0s with permissible places for the two 1's marked x
x 0 x 0 x 0 x 0 x 0 x 0 x
the 1's can be placed in 7c2 ways
when there are five 0's, a similar diagram will give 6c3 places for the three 1's
if you have more than four zeos, a permissible arrangement isn't possible
so starting for zero 0's, the permissible ways are
9c0 + 8c1 + 7c2 + 6c3 + 5c4
in your men-women case, each person is distinct
whereas here all 1's or 0's are identical.
moreover, there are 5 possible cases here, whereas
the man-woma case is only one
let there be six 0s with permissible places for the two 1's marked x
x 0 x 0 x 0 x 0 x 0 x 0 x
the 1's can be placed in 7c2 ways
when there are five 0's, a similar diagram will give 6c3 places for the three 1's
if you have more than four zeos, a permissible arrangement isn't possible
so starting for zero 0's, the permissible ways are
9c0 + 8c1 + 7c2 + 6c3 + 5c4
in your men-women case, each person is distinct
whereas here all 1's or 0's are identical.
moreover, there are 5 possible cases here, whereas
the man-woma case is only one
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i guess you're right! but i will prefer this way: 4!*9P5.