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Numerical Ability
Permutation and Combination
In how many ways 11women and 5 men can be suffeled to get 11 members with atleast 3men?
Read Solution (Total 26)
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- answer is 3762
atleast 3 men means we need to incude 3 or more men for every group of 11 members.
5c3*11c8+5c4*11c7+5c5*11c6 - 9 years agoHelpfull: Yes(7) No(0)
- 5c3*11c8+5c4*11c7+5c5*11c6
- 9 years agoHelpfull: Yes(2) No(0)
- 5c3*11c8+5c4*11c7+5c5*11C6=3762 ways
- 9 years agoHelpfull: Yes(2) No(0)
- sorry... that should be 10 * 165 = 1650
- 9 years agoHelpfull: Yes(1) No(0)
- 16C11 -( 5C0 X 11C11) - (5C1 X 11C10) - ( 5C2 X 11C9) = ANS
first choose all possible combinations (ie. 11 people out of 11+5 people) then subtracting unfavorable cases (ie in which 0men&11 women were selected or 1 men&10 women were selected or 2 men and 9 women were selected)
On solving answer comes to be 3762
- 9 years agoHelpfull: Yes(1) No(0)
- 8 women+3 men=11 member. (out of 11 women and 5 men)
7 women +4 men =11 member. (out of 11 women and 5 men)
6 women+5 men=11 member. (out of 11 women and 5 men)
so,(11c8*5c3)+(11c7*5c4)+(11c6*5c5)=3762 - 9 years agoHelpfull: Yes(1) No(0)
- 5c3 + 11c8 = 10 + 165
175 - 9 years agoHelpfull: Yes(0) No(0)
- total no of ways=5c3*11c8+5c4*11c7+5c5*11c6=3762
- 9 years agoHelpfull: Yes(0) No(0)
- that will be 5c3 * 11c8 + 5c4 * 11c7 + 5c5 * 11c6
- 9 years agoHelpfull: Yes(0) No(0)
- 16c5- (11c11+(11c10*5c1)+(11c4*5c2))
- 9 years agoHelpfull: Yes(0) No(0)
- plz explain clearly
- 9 years agoHelpfull: Yes(0) No(0)
- we need 11 members with ATLEAST 3 men so..
(11c11+5c0)+(11c10+5c1)+(11c9+5c2)+(11c8+5c3)=(1+1)+(11+5)+(55+10)+(11*15+10)=258...i think - 9 years agoHelpfull: Yes(0) No(0)
- answer is 3762
atleast 3 men means we need to incude 3 or more men for every group of 11 members.
5c3*11c8+5c4*11c7+5c5*11c6 - 9 years agoHelpfull: Yes(0) No(0)
- we need 11 members with ATLEAST 3 men so..
(11c11*5c0)+(11c10*5c1)+(11c9*5c2)+(11c8*5c3)=(1*1)+(11*5)+(55*10)+(11*15*10)=2266...i think
- 9 years agoHelpfull: Yes(0) No(0)
- 11c11+ 11c10*5c1 + 11c9*5c2 + 11c8*5c3
- 9 years agoHelpfull: Yes(0) No(0)
- we can select atleast 3 men fromb5 men so 5c3
and we want 11 members so remaining are 8 we take from women as 11c8
then finally 11c8*5c3=19800
- 9 years agoHelpfull: Yes(0) No(0)
- in group there has to be 3 men -for this we can select 3 men from 5 in 5c3 ways and remaining 8 from 11 women in 11c3 ways
therefore total ways:- 5c3*11c3 - 9 years agoHelpfull: Yes(0) No(0)
- 5C3 x 11C8 + 5C4 x 11C7 + 5C5 x 11C6
= 3762 - 9 years agoHelpfull: Yes(0) No(0)
- 11c8*5c3 + 11c7*5c4 + 11c6=3762
- 9 years agoHelpfull: Yes(0) No(0)
- 11 women 5 men ways of selection
8 3 11C8 * 5C3
7 4 11C7 * 5*4
6 5 11C6 * 5C5
Hence no. of selecting is
(11C8 * 5C3)+(11C7 * 5C4)+(11C6 * 5C5) =3762 WAYS - 9 years agoHelpfull: Yes(0) No(0)
- 5c3 * 11c8=1650
- 9 years agoHelpfull: Yes(0) No(0)
- 16c11- (11c11 + 11c10*5c1 + 11c9*5c2)
- 9 years agoHelpfull: Yes(0) No(0)
- 5c3*11c8+5c4*11c7+5c5*11c6
=3762 - 9 years agoHelpfull: Yes(0) No(0)
- 2442 is theans
- 9 years agoHelpfull: Yes(0) No(0)
- 5c3*11c8+5c4*11c7+5c5*11c6
- 9 years agoHelpfull: Yes(0) No(0)
- 5c3*11c8+5c4*11c7+5c5*11c6=3762
- 9 years agoHelpfull: Yes(0) No(0)
TCS Other Question
Que. Given A+B+C+D+E= FG. If FG is Maximum as possible and A,B,C,D,E,F,G all are distinct digits(0-9) then what is the value of G.
a) 1
b) 2
c) 3
d) 4
e) 5
a three digit number, sum of the digit is 17, the sum of square of digits is 109, if it is subtracted wit 495 the number gets reversed, find the three digit number.
A) 683
B) 863
C) 368
D) 686