1. How many permutations of the letters ABCDEFG contain the string BCD?

2.How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other

Please tell me the solution of how to get it .. TNX

Guest Mar 2, 2015

#3**+5 **

2.How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other

ok, here is the line

?M?M?M?M?M?M?M?M?

The M represents a man and the ? represents where a woman could stand

there are 8! permutations for the men (8P8)

Now there are 9 places where the 5 women could stand so that is 9P5

Put them together and you have 8!*9P5 permutations

$${\mathtt{8}}{!}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{9}}{!}}{({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{609\,638\,400}}$$

I think that is correct.

Melody
Mar 3, 2015

#1**+5 **

BCD and AEFG I am assuming that the BCD must stay together in that order.

The aefg can be in any order so there are 4! permutations there.

Now for each of those permutations BCD can be before or after the first, or before or after the last or in the middle that is 5 different positions.

So the answer is 4!*5

$${\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{5}} = {\mathtt{120}}$$

Melody
Mar 2, 2015

#2**0 **

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CPhill
Mar 2, 2015

#3**+5 **

Best Answer

2.How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other

ok, here is the line

?M?M?M?M?M?M?M?M?

The M represents a man and the ? represents where a woman could stand

there are 8! permutations for the men (8P8)

Now there are 9 places where the 5 women could stand so that is 9P5

Put them together and you have 8!*9P5 permutations

$${\mathtt{8}}{!}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{9}}{!}}{({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{609\,638\,400}}$$

I think that is correct.

Melody
Mar 3, 2015