Latest Elitmus Aptitude Question SOLUTION: asked on 13/09/2015 There are 21 students in a class where boys number exceeds girls no by 1 . A quiz group of 4 students has to be organized where all 4 should not be from same ge

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(#M40153682) ELITMUS Numerical Ability question permutation and combination Keep an EYE

asked on 13/09/2015
There are 21 students in a class where boys number exceeds girls no by 1 .
A quiz group of 4 students has to be organized where all 4 should not be from same gender . Then how many ways quiz groups can be created
Options I dnt remember but option B was like 5445

Asked In Elitmus Chetan Anand (8 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

raviranjan  (8 years ago)

b+g=21
b-g=1
----------
2g=20
g=10,
b=11
So,Total Combination are
bbbg,bbgg,bggg
(11c3 * 10c1+11c2*10c2+11c1*10c3)
1650+2475+1320=5445

Is this solution Helpfull? Yes (53) | No   

Ajitesh rai  (8 years ago)

Ans = Total no of ways - No of ways of groups of same gender

= 21C 4 - { 11C 4 + 10C 4 }

= 5445

Is this solution Helpfull? Yes (23) | No (1)  

Vikas Jha  (8 years ago)

b) 5445
x= no of boys, y= no of girls
x+y=21....................................(1)
x-y=1 ....................................(2)
Solve these x=11,y=10
Now Group of 4 where all 4 are not from the same gender can be..
1) 1Boy3Girls= 11c1*10c3
2) 2Boys2Girls= 11c2*10c2
3) 3Boys1Girl= 11c3*10c1
Hence total ways will be = 11c1*10c3+11c2*10c2+11c3*10c1=1320+2475+1650=5445
So, If You have guessed b) It's Right.. :)

Is this solution Helpfull? Yes (6) | No   

rajat gupta  (8 years ago)

total girls are 10 and boys are 11 so no of ways are 10c1*11c3+10c2*11c2+10c3*11c1
which is 1650+2475+1320=5445 &b IS THE ANSWER

Is this solution Helpfull? Yes (1) | No   

Dhwrwm Basumatary  (8 years ago)

as per question, 11 boys and 10 girls ...
Total number of ways of groups created by 21 peolpes = 21c4=5985
# of ways in creating grp consiting of only men = 11c4=330
# of ways in creating grp consiting of only women = 10c4=210
The reqd # of ways = 5985-330-210=5445

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Mayank Aditya  (8 years ago)

ans-5445
soln-
boy=11
girl=10

no of ways =[3boy1girl]+[2boy2girl]+[1boy3girl]
= [11C3*10C1+11C2*10C2]+[11C1*10C3]
=5445

Is this solution Helpfull? Yes | No   

avinash surama  (8 years ago)

11 boys and 10 girls ...
11c3 * 10c1 + 11c2*10c2 + 11c1*10c3 = 8470

Is this solution Helpfull? Yes | No (2)  

dipanjan ghosh  (8 years ago)

Answer is probably option b 5445.
21= x + (x-1), where x is no. of boys.
Hence 11 boys and 10 girls.
Now, (11C3 * 10C1) + (11C2*10C2) + (11C1 * 10C3) =5445.

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subham kumar  (8 years ago)

5445
given no.of boys is 1 more than girl .
let no. of boys be n
then no of girl is n-1
a/q
n+n-1=21
n=11
now three cases arises
case 1: selecting 2 boys and 2 girls
11c2*10c2
case 2: selecting 3 boys and 1 girls
11c3*10c1
case 3: selecting 1 boys and 3 girls
11c1*10c3
as the above cases is mutually exclusive so
total no of selection i=11c2*10c2+11c3*10c1+11c1*10c3=5445

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priyanka  (8 years ago)

Hi Chetan.. You got your result for 13/09/2015 exam?

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Dhwrwm Basumatary  (8 years ago)

Boys=11, girls =10;
21 students can be arranged in 21c4 ways = 5985 ways
11 boys can ...... in 11c4ways = 330 ways
10 girls can ...... in 10c4=210 ways
There a group without all members of same gender = (total ways)-(only boys)-(only girls)
=5985-330-210=5445

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nitin  (8 years ago)

Proper explanation From this concept you can solve any case

Total 6 numbers are there 1,2,3,4,5,6 among them we have to make 5 digits
04= rest remaining 4 so they have to fit in 3 slots!! so ways is 4!
20= rest remaining 4 so they have to fit in 3 slots!! so ways is 4!
40= rest remaining 4 so they have to fit in 3 slots!! so ways is 4!
12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
32=12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
24=12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
52=12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
therefore add all cases=4!+4!+4!+18+18+18+18=144 ans.

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