[Math] How to arrange 15 women and 15 men so no two females are seated next to each other

permutationsprobability

To a certain conference, each firm can send two employee representatives, on the condition that one of them is a male and the other a female. If 15 firms were represented in this conference, what is the probability that no two females are seated next to each other? (Assume that all 30 members are seated in a single row of seats)
A] 1/2
B] 14!/30!
C] 15!/30!
D] 28/30!

For above question I am trying following: 15 men can be arranged in 15! ways. There are 16 spots among men (-M-M-…..-M-). 16 spots can be taken by 15 women in 16! ways. My answer is (15!*16!)/30!. But it's not listed in answer choices. Am I missing something or answers are incorrect?

My approach is similar to approach in question in below link.
How many ways are there for 10 women and six men to stand in a line

Best Answer

$\bf{My\; Solution::}$ If we can represent man by $\bf{M}$ and Women by $\bf{W}$, Then we use Gap Method.

So Arrangements as $$_M_M_M_M_M_M_M_M_M_M_M_M_M_M_M_$$

Above $15$ man can be arrange as $15!$ (Here Man and women all are Different)

Now We can Arrange $15$ Women in These $16$ Gap, Which can be done by $\displaystyle \binom{16}{15}\times 15! = 16!$

So Total no. of Arrangement in which no $2$ woman sit together is $\displaystyle 15! \times 16!$

And Total Probability $\displaystyle = \frac{\bf{Favourable \; cases}}{\bf{Total \; Cases}} = \frac{15!\times 16!}{30!}$

So I Think Your answer is Right.

Related Solutions

[Math] How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other

For this type of problem first consider the position of the men, then the position of the women.In general first consider the the group with more people.

How many possible ways are there to arrange eight men in a row? It will be $ ^8P_8 = 8! = 40320$

Now as no two women stand next to each other,we can imagine the situation as

                   * M * M * M * M * M * M * M * M *

Hence, we need to find how many ways we can arrange $5$ women in the $9$ possible (as shown above) places,this is actually $ ^9P_5 = 9*8*7*6*5 = 15120$

Now applying the fundamental law of counting (precisely product rule), total number of possible arrangements satisfying both constraints is: $15120 * 40320 = 609638400$ which is your required/desired answer.

[Math] Find the number of seating arrangements at a round table of three single men, two single women, and two families

For $1$, each family forms a block. There are $7!/7$ inequivalent ways to arrange the $7$ men, women and blocks. With the single women also forming a block, there are $6!/6$ inequivalent ways. In the first case, there are $3!^2$ internal permutations for the families, and in the second case there are $3!^2$ internal permutations for the families and $2$ for the single women. Thus the total number of inequivalent admissible arrangements is

$$ (3!^2)7!/7-2(3!^2)6!/6=17280\;. $$

If the seats are numbered and each child sits between their parents, there are $11\cdot7!/7$ arrangements of the men, women and family blocks, and $2^2$ internal arrangements for the two families, yielding

$$ 2^2\cdot11\cdot7!/7=31680 $$

different arrangements.

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Related Solutions

[Math] How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other

[Math] Find the number of seating arrangements at a round table of three single men, two single women, and two families

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