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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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A lady gives a dinner party to six guests. The number of ways in which...
The lady has ten friends. Two of these friends will not agree to go to the party together. So, let us consider the following case: neither of the two friends will attend the party. So, the number of ways six guests can be invited in this case will be equal to the number of ways to choose 6 friends out of the remaining 8 friends. So, we have
Therefore, there are 28 ways to invite 6 friends when the two friends will not be going to the party.
Now, we will consider the next case, which is that one of the two friends decides to attend the party. So, if either of the two decide to attend the party, then the rest 5 guests will be chosen from the remaining 8 friends. Therefore, if one friend decides to attend the party, the number of ways to invite the other 5 guests is 8C5 . Similarly, if the other friend decides to attend the party, the number of ways to invite the other 5 guests is 8C5 . Therefore, the number of ways to invite the guests if either, if the two friends is going to the party, is the following,
Now, we will consider the next case, which is that one of the two friends decides to attend the party. So, if either of the two decide to attend the party, then the rest 5 guests will be chosen from the remaining 8 friends. Therefore, if one friend decides to attend the party, the number of ways to invite the other 5 guests is 8C5 . Similarly, if the other friend decides to attend the party, the number of ways to invite the other 5 guests is 8C5 . Therefore, the number of ways to invite the guests if either, if the two friends is going to the party, is the following,
Therefore, the total number of ways to invite six guests out of ten with the given conditions is 28+112 = 140.
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A lady gives a dinner party to six guests. The number of ways in which...
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.
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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