Free Practice Question
PS · Problem Solving
705-805
Combinations
In how many ways 5 men and 4 women can be arranged in a line so that no two women are besides each other?
Answer Choices
Correct answer marked belowA
120
B
360
C
600
43,200
Correct
E
46,900
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Explanation
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Official Explanation
Official
Solution provided by sandeep800 is correct.
Consider this: 5 men can be arranged in 5! number of ways.
Consider one particular arrangement of these men and empty slots as follows: *m*m*m*m*m*.
Now, if we place 4 women in either of 4 slots out of 6 then no two women will be together --> is the # of ways to choose in which 4 slots out of 6 these women will be placed and 4! is # of arrangements of them in these slots.…
Consider this: 5 men can be arranged in 5! number of ways.
Consider one particular arrangement of these men and empty slots as follows: *m*m*m*m*m*.
Now, if we place 4 women in either of 4 slots out of 6 then no two women will be together --> is the # of ways to choose in which 4 slots out of 6 these women will be placed and 4! is # of arrangements of them in these slots.…
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What it tests
Whether you can count arrangements and selections correctly — permutations vs. combinations.
Common trap
Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.
Details
Difficulty
705-805
Type
PS
Category