Each of two women and three men is to occupy one chair out of eight chairs, each of which is numbered from one to right. First, women are to occupy any two chairs from those numbered one to four; and then the three men would occupy any three chairs out of

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Q: 37 (IAS/2006)
Each of two women and three men is to occupy one chair out of eight chairs, each of which is numbered from one to right. First, women are to occupy any two chairs from those numbered one to four; and then the three men would occupy any three chairs out of the remaining three men would occupy any three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done?

question_subject: 

Logic/Reasoning

question_exam: 

IAS

stats: 

0,8,8,3,4,8,1

keywords: 

{'chairs': [0, 0, 1, 0], 'maximum number': [1, 0, 1, 2], 'chair': [0, 0, 2, 0], 'women': [9, 8, 22, 46], 'men': [4, 3, 12, 9], 'different ways': [0, 0, 5, 0]}

This question involves the concept of permutation and combinations. We have two women who have to occupy any two chairs from the first four, the number of ways this can happen is 4P2 (Permutations of 4 items taken 2 at a time) that equals to 12. Then, the three men have to occupy any three of the remaining six chairs. The number of ways to do this is 6P3 (Permutations of 6 items taken 3 at a time) which equals 120. The maximum number of different ways is gotten by multiplying these results: 12*120 = 1440.

Option 1: 40 is incorrect as it doesn`t account for the different permutations correctly as mentioned.

Option 2: 132 also incorrectly calculates the permutations.

Option 3: 1440 is the correct answer because it correctly calculates the permutations of the chairs for both the women and men.

Option 4: 3660, like options 1 and 2, miscalculates the permutations.

Hence, option 3, 1440, is the maximum number of different ways to seat the women and men.

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