Given 10 people P1,P2,...,P10,how many 6-member
teams can be formed if
(a) both P2,and p4 must be chosen?
(b) at most one of P2,P4 can be chosen?
Answers
Given : 10 people P1,P2,...,P10
6-member teams to be formed
To Find : number of Ways
(a) both P2,and p4 must be chosen?
(b) at most one of P2,P4 can be chosen?
Solution:
10 people P1,P2,...,P10,
6 member teams
P2 and P4 must be chosen
Hence remaining 4 has to be chosen from remaining 8
4 can be chosen from 8 in
⁸C₄ ways
= 8!/4!.4!
= 70
in 70 ways 6 members team can be formed if both P2,and p4 must be chosen
at most one of P2, P4 can be chosen
if P2 is chosen - then remaining 5 to be selected from 8 other than P2 & P4.
P2 chosen P4 not chosen = ⁸C₅ = 56 way
if P4 is chosen - then remaining 5 to be selected from 8 other than P2 & P4.
P4 chosen P2 not chosen = ⁸C₅ = 56 way
if P2 & P4 not chosen then all 6 to be selected from 8 other than P2 & P4.
P2 & P4 not chosen = ⁸C₆ = 28 Way
56 + 56 + 28 = 140 Ways
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