In a class of 21 students we have to organise a quiz competition in which we have to selevt at

grls(G)=10,boys(B)=11
1 G & 3 B=>10c1*11c3=1650
3 G & 1 G=>10c3*11c1=1320
1650+1320=2970
9 years agoHelpfull: Yes(29) No(0)

g=10,b=11 ans=2970
9 years agoHelpfull: Yes(6) No(0)
given that boys are 1more than girls.

so B=G+1
=> B-G=1

now we have total no of students 21
i.e B+G=21

solve these two equations

B-G=1
B+G=21

so we get B=11, G=10

no we have to choose 4 students from 21

21c4

for 1 G,3B => 10c1* 11C3= 1650
for 3G,1B => 10C3* 11C1= 1320

1650+1320= 2970
9 years agoHelpfull: Yes(5) No(0)
Solution :
as per the question we have 11 boys and 10 girls
Case-1 when we have to select 1girl and 3 boys
this can be done
11C3*10C1=1320

Case2 when we have to select 1 boy and 3 girls
this can be done
11C1*10C3=1650

Adding both the cases we get
1320+1650= 2970 ways
9 years agoHelpfull: Yes(3) No(0)
Girl=10, Boys=11 (because Boys are 1 more than the girls in the class)
(10C1 * 11C3) + (10C3 + 11C1)
=1650 + 1320=2970
8 years agoHelpfull: Yes(1) No(0)
Can u plz provide the complete solution
9 years agoHelpfull: Yes(0) No(1)
2970. is the ans.
9 years agoHelpfull: Yes(0) No(0)
girls =10 ,boys=11
10c1*11c3+11c1*10c3=2970
9 years agoHelpfull: Yes(0) No(0)
10c1*11c3 + 11c1*10c3=2970
9 years agoHelpfull: Yes(0) No(0)
10*11c3+11*10c3=2970
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B+(B+1)=21
=>B=11
=>G=10
Req no of ways=(10c1*11c3)+(11c1*10c3)
=1650+1320
=2970 ways
9 years agoHelpfull: Yes(0) No(1)
The last two digits will be filled in 8 ways i.e 04,08,12,20,24,32,40,52,20. and the rest will be 3 *3 9 ways thus total 9*8 =72 ways
8 years agoHelpfull: Yes(0) No(0)

In a class of 21 students ,we have to organise a quiz competition in which we have to selevt at leat 4 students such that (1 girl and 3 boys) or (1 boy 3 girls.).Boys are 1 more than the girls in the class?