Answer
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Hint: We use the principle of permutations and combinations to find the number of ways in which we can arrange men and women (make them sit in a row) such that we can meet the above constraint so that the women occupy the even places.
Complete step-by-step answer:
First, before we solve this problem, we try to understand how we can arrange 5 men and 4 women in a row without any constraints. To count the number of combinations for this case, we basically have 9 people (we have no constraints so there is no difference between men and women while counting the number of combinations), we have the combinations as 9!
However, the required number of cases with constraints, clearly we would have a lesser number of cases than 9! arrangements. Thus, to give an idea about the current problem, we have,
M W M W M W M W M (Basically, our arrangement would look like this since women would occupy even places.)
Thus, the places for the women are fixed, thus the total number of arrangements is 4! (which is $^{4}{{P}_{4}}$).
Similarly, due to the women’s places getting fixed, men’s places automatically fixed, thus the total number of arrangements is 5! (which is $^{5}{{P}_{5}}$).
Complete step-by-step answer:
First, before we solve this problem, we try to understand how we can arrange 5 men and 4 women in a row without any constraints. To count the number of combinations for this case, we basically have 9 people (we have no constraints so there is no difference between men and women while counting the number of combinations), we have the combinations as 9!
However, the required number of cases with constraints, clearly we would have a lesser number of cases than 9! arrangements. Thus, to give an idea about the current problem, we have,
M W M W M W M W M (Basically, our arrangement would look like this since women would occupy even places.)
Thus, the places for the women are fixed, thus the total number of arrangements is 4! (which is $^{4}{{P}_{4}}$).
Similarly, due to the women’s places getting fixed, men’s places automatically fixed, thus the total number of arrangements is 5! (which is $^{5}{{P}_{5}}$).
Now, combining these results, we have a total number of combinations of 4!$\times $5!=2880.
Note: While solving the questions related to permutations and combinations, it is important to know the techniques of counting using the permutation and combinations notation. This is because for small cases (suppose case of 3 people) we can actually count by hand with enlisting the cases (that is for 3 people, we can actually enlist 3! =6 cases by hand), however, the same cannot be done for larger cases (example 7 people where cases would be 7! =5040).
Note: While solving the questions related to permutations and combinations, it is important to know the techniques of counting using the permutation and combinations notation. This is because for small cases (suppose case of 3 people) we can actually count by hand with enlisting the cases (that is for 3 people, we can actually enlist 3! =6 cases by hand), however, the same cannot be done for larger cases (example 7 people where cases would be 7! =5040).
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