Lesson Men and women standing in line

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Men and women standing in line

Problem 1

Eight men and five women are standing in a line.
In how many arrangements will no two women be standing next to one another?

Solution

Imagine that the 8 men are standing in line with intervals (gaps) between them, as shown in the Figure below.


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


(" * "s  show intervals, or gaps). You see 7 stars between "M"s and two stars before and after the M-line; in all, 7+2 = 9 stars.

These stars are the only places for women: Each woman can occupy one and only one position at the star place.


So, we can position 8 men by 8! = 8*7*6*5*4*3*2*1 = 40320 ways.


And we can place first  woman in any of nine (*)-positions;

                 second woman in any of remaining eight (*)-positions;

                 third woman  in any of remaining seven positions;

                 fourth . . . 

             and fifth . . . 


Thus, in all, there are  8!*9*8*7*6*5 = 40320*9*8*7*6*5 = 609638400 differennt arrangements satisfying given conditions.    ANSWER

Problem 2

Eight women and five men are standing in a line.
In how many arrangements will no two men be standing next to one another?

Solution

Imagine that the 8 women are standing in line with intervals (gaps) between them, as shown in the Figure below.


                     * W * W * W * W * W * W * W * W *


(" * "s  show intervals, or gaps). You see 7 stars between "W"s and two stars before and after the W-line; in all, 7+2 = 9 stars.

These stars are the only places for men: Each man can occupy one and only one position at the star place.


So, we can position 8 women by 8! = 8*7*6*5*4*3*2*1 = 40320 ways.


And we can place first  man in any of nine (*)-positions;

                 second man in any of remaining eight (*)-positions;

                 third  man in any of remaining seven positions;

                 fourth . . . 

             and fifth . . . 


Thus, in all, there are  8!*9*8*7*6*5 = 40320*9*8*7*6*5 = 609638400 differennt arrangements satisfying given restrictions.    ANSWER

Surely, these two problems are TWINS.   I placed them both together to highlight the major idea of the solution.

            We first arrange a greater amount of items in a line.
            Then we place the other sort of items in the gaps of this line.

My other lessons on Permutations and Combinations in this site are
    - Introduction to Permutations
    - PROOF of the formula on the number of Permutations
    - Simple and simplest problems on permutations
    - Special type permutations problems
    - Problems on Permutations with restrictions
    - Math circle level problem on Permutations
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - Problems on Combinations with restrictions
    - Math circle level problem on Combinations
    - Arranging elements of sets containing indistinguishable elements
    - Persons sitting around a cicular table
    - Combinatoric problems for entities other than permutations and combinations
    - Fundamental counting principle problems
    - Miscellaneous problems on permutations, combinations and other combinatoric entities
    - In how many ways N distinguishable objects can be distributed among n different boxes ?
    - Stars and bars method for Combinatorics problems
    - Nice recreational problems on permutations
    - OVERVIEW of lessons on Permutations and Combinations

Use this file/link ALGEBRA-II - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-II.

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