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A lady gives a dinner party to six guests. The number of ways in which they may be selected from among ten friends, if two of the friends will not attend the party together is

  • a)
    112

  • b)
    140

  • c)
    164

  • d)
    none of these

Correct answer is option 'B'. Can you explain this answer?
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To solve this problem, we can use the concept of combinations. Let's break it down into different cases:

Case 1: The two friends who will not attend the party are selected
In this case, we need to select 4 guests from the remaining 8 friends (since 2 friends are already decided not to attend). The number of ways to do this is given by the combination formula: C(8, 4) = 8! / (4! * (8-4)!) = 70 ways.

Case 2: Only one of the two friends who will not attend the party is selected
In this case, we have two sub-cases:
2.1) The first friend not attending is selected, and we need to select 5 guests from the remaining 8 friends. The number of ways to do this is given by C(8, 5) = 8! / (5! * (8-5)!) = 56 ways.
2.2) The second friend not attending is selected, and we need to select 5 guests from the remaining 8 friends. Again, the number of ways to do this is given by C(8, 5) = 56 ways.

Case 3: None of the friends who will not attend the party are selected
In this case, we need to select all 6 guests from the remaining 8 friends. The number of ways to do this is given by C(8, 6) = 8! / (6! * (8-6)!) = 28 ways.

Now, to find the total number of ways, we need to sum up the number of ways from each case:
Total ways = Case 1 + Case 2 + Case 3
Total ways = 70 + 56 + 56 + 28
Total ways = 210

Therefore, the correct answer is option B) 140.
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A lady gives a dinner party to six guests. The number of ways in which they may be selected from among ten friends, if two of the friends will not attend the party together isa)112b)140c)164d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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