In how many ways can 5 members be selected out of 10 members,so that any two particular members must always be excluded?
a- 52
b-56
c-60
d-64

• b-56. since 2 particular members are always excluded then you are actually choosing from 8 members. and the number of ways 5 members are selected is equal to 8C5 which is equal to 56.
• 8 years agoHelpfull: Yes(28) No(0)
• two particular members must be excluded so member=10-2=8
  so the answer will be = 8! / 5! (8-5)!=8*7*6*5!/5!*3!=8*7*6/3*2*1=8*7=56
• 8 years agoHelpfull: Yes(8) No(0)
• You have to choose 5 members from 8 .because 2 people always excluded. Therefore 8C5 is equal to 56.
• 8 years agoHelpfull: Yes(5) No(0)
• ans is 56.couple of people selected.so remai9ning 8.8c5-ans is 56
• 8 years agoHelpfull: Yes(1) No(0)
• as 2 members are always excluded then we left with only 8 members and from that 5 members are selected so combination is use -> 8C5= 56
• 7 years agoHelpfull: Yes(1) No(0)
• it will be 8!/(5!*3!)=56
• 7 years agoHelpfull: Yes(0) No(0)
• exclude 2 out of 10 i.e.., 8
  select 5 out of 8
  8c5 = 56
• 7 years agoHelpfull: Yes(0) No(0)
• b-56
  Since two members should be excluded total members are 8.
  from eight we can select 5 as 8C5 which is 56.
• 7 years agoHelpfull: Yes(0) No(0)
• 2 ppl are excluded always ie 8C5 = 56
• 7 years agoHelpfull: Yes(0) No(0)
• 8c2 =56
  Because 2 people are excluded in 10
• 7 years agoHelpfull: Yes(0) No(0)
• 2 r always excluded so it is 8C5 =56
• 7 years agoHelpfull: Yes(0) No(0)
• Two members always be excluded means 8 members should be availablle ..So we can select 5 members from 8 members (ie) 8c5=56
• 7 years agoHelpfull: Yes(0) No(0)
• B. 56
  since, 2 members must be excluded so remaining member is only 8
  so,
  8C6 = 8! / 3! = 56
• 7 years agoHelpfull: Yes(0) No(0)
• 2 mans always excluded then there are 8 members available, then select 5 members out of 8 .
  so the possible ways are 8c5.
  
• 7 years agoHelpfull: Yes(0) No(0)
• 8 c 5=56.
  
• 7 years agoHelpfull: Yes(0) No(0)