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A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two particular friends A and B will not attend the party together is
- a)56
- b)126
- c)91
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A lady gives a dinner party to 5 guests to be selected from nine frien...
Let the 9 friends be A, B, C, D, E, F, G, H and i respectively.
A and B do not attend the party together.
Total number of ways to select 5 from 7 = 7C5
= (7*6)/(2*1)
= 42/2
= 21 ways
Either of A or B is selected for party, then number of ways = 2C1*7C4
= (2*1)*(7*6*5)/(3*2*1)
= 420/6
= 70 ways
Total number of ways = 21 + 70
= 91 ways
Most Upvoted Answer
A lady gives a dinner party to 5 guests to be selected from nine frien...
Given: A lady gives a dinner party to 5 guests to be selected from nine friends. Two particular friends A and B will not attend the party together.
To find: The number of ways of forming the party of 5.
Solution:
Let's consider two cases:
Case 1: When A is selected
In this case, B cannot be selected. So we need to select 4 guests from the remaining 7 friends.
Number of ways of selecting 4 guests from 7 friends = 7C4 = 35
Case 2: When A is not selected
In this case, B can be selected. So we need to select 5 guests from the remaining 7 friends (excluding A and B).
Number of ways of selecting 5 guests from 7 friends = 7C5 = 21
Total number of ways of forming the party of 5 = Number of ways of forming a party of 5 when A is selected + Number of ways of forming a party of 5 when A is not selected and B can attend
= 35 + 21
= 56
But we need to exclude the case where A and B attend together.
Number of ways of selecting 3 guests from the remaining 7 friends (excluding A and B) when A and B attend together = 7C3 = 35
So, the number of ways of forming the party of 5, given that A and B will not attend the party together = Total number of ways of forming the party of 5 - Number of ways of forming a party of 5 when A and B attend together
= 56 - 35
= 21
Therefore, the correct option is (c) 91.
To find: The number of ways of forming the party of 5.
Solution:
Let's consider two cases:
Case 1: When A is selected
In this case, B cannot be selected. So we need to select 4 guests from the remaining 7 friends.
Number of ways of selecting 4 guests from 7 friends = 7C4 = 35
Case 2: When A is not selected
In this case, B can be selected. So we need to select 5 guests from the remaining 7 friends (excluding A and B).
Number of ways of selecting 5 guests from 7 friends = 7C5 = 21
Total number of ways of forming the party of 5 = Number of ways of forming a party of 5 when A is selected + Number of ways of forming a party of 5 when A is not selected and B can attend
= 35 + 21
= 56
But we need to exclude the case where A and B attend together.
Number of ways of selecting 3 guests from the remaining 7 friends (excluding A and B) when A and B attend together = 7C3 = 35
So, the number of ways of forming the party of 5, given that A and B will not attend the party together = Total number of ways of forming the party of 5 - Number of ways of forming a party of 5 when A and B attend together
= 56 - 35
= 21
Therefore, the correct option is (c) 91.
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A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two particular friends A and B will not attend the party together isa)56b)126c)91d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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