A lady gives dinner party to five guests to be selected from 9 friends .the number of ways of forming the party of 5,given that two of the friends will not attend the party together is
Answers
Answered by
36
Solution :-
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
Answer.
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
Answer.
Answered by
3
Answer:
91
Step-by-step explanation:
Scene1: among 9 friends 5 guests can be selected in 9C5 = 126 ways.
Scene 2: two friend will attend party together and therefore, they are occupying two invitations out of 5 already. So 3 invitees are left i.e From rests 7 friends now you have to choose 3 friend to invite. Simply in 7C3 = 35 ways.
scene3: those two friends wont attend party together i.e scene2 will be eliminated from scene 1 ——> 126 - 35 = 91 ways .............Ans.
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