Latest Elitmus Aptitude Question SOLUTION: There are nine politicians A, B, C, D, E, F, G, H, I who have to become member of one of the 3 commitees namely I, II, III. Each politician should become member of atleast 1 comm

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(#M40030095) ELITMUS question Keep an EYE

There are nine politicians A, B, C, D, E, F, G, H, I who have to become member of one of the 3 commitees namely I, II, III. Each politician should become member of

atleast 1 commitee & only in 1 commitee. I has exactly 1 more than II, III may or may not contain any members

A, B, C can not become members of I
D, E, F can not become members of II
G, H, I can not become members of III

1. If II has exactly 3 members, then in how many ways different commitees can form.

2. If II has exactly 3 members, then in how many ways III can have members.

Asked In Elitmus Sagar Dasari (10 years ago)

Unsolved Read Solution (8)

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Harshit  (10 years ago)

1. if 2nd have ABC then 3C3(II)*6C4(I)*2C2(III)= 15
if 2nd have any 2 from GHI(beacuse it can't take all three)and any 1 from ABC then 3C2*3C1(II)*4C4(I)*2C2(III)=9
if 2nd have any 1 from GHI and 2 from ABC then 3C1*3C2(II)*5C4(I)*2C2(III)=45
15+9+45=69 ways.
2.if ABC are fixed to 2nd then if we first assign to III then it can choose any 2 member from remaining 6 so 6C2=15
if 2nd have 2 member from GHI then III can take only reamining 2 values from ABC(any two from ABC as 2nd can't take value from DEF so it will take 2 from GHI and 1 from ABC) 2C2=1
if 2nd have 1 member from GHI then III can take any 1 from DEFGHI and 1 From remaining 1 in ABC so 5c1*1c1=5
then 15+1+5=21 way

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Shreesh Tripathi  (10 years ago)

qus 1- exactly 3 in II i.e 4 in I and hence 2 in III
this in I- 6 members are allowed except A,B,C so 6*5*4*3
in II-remaining 5*4*3 ways
and III -remaining 2 in 2 ways

Qus 2-similiraly in III 2 ways

Is this solution Helpfull? Yes (2) | No (7)  

manish kumar sinha  (10 years ago)

bhai iska koi correct solution do

Is this solution Helpfull? Yes (2) | No   

ASTHA Tripathi  (10 years ago)

1.14,first fix ghi in 2nd gp and then fix abc in 2nd gp

2.14

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komal pandhariwal  (10 years ago)

I II III
_ _ _ _ _ _ _ _ _ =9 seats

1)ABC a) A B C _ _ (1 way)
b) A B _ C _ (3c1 ways)
c) A _ _ B C (3c1 ways)

2)DEF a) D _ _ _ E F (3c1 ways)
b) D E _ _ F _ (3c1 ways)
c) D E F _ _ _ (1 way)

3)GHI a) G H I _ (1 way)
b) G H _ _ I _ _ (3c1 ways)
c) G _ _ _ H I _ (3c1 ways)

1)by combining a, b, c of 1, 2, 3 we get (1*3*3)+(3*3*1)+(1*3*3)=27
2)we check where III is filled in the table accordingly (3*3)+(3)+(3)=15

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vimal  (10 years ago)

what is r8b sol yrr

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puneet khandelwal  (10 years ago)

1.I-4MEMBERS;II-3MEMBERS;III-2MEMBERS
I CANT HAVE ABC SO NO OF WAYS OF CHOOSING 4 MEMBERS OUT OF 6 MEMBERS(DEFGHI)IS6C4=15;
NOW 15 VALID WAYS ARE
I II III NO OF WAYS OF DOING AS PER QUES
DEFG 3
FGHI 1
EGHI 1
DGHI 1
GDEF 3
HDEF 3
IDEF 3
DEGH 3
DEHI 3
EFGH 1
EFGI 3
EFHI 3
DFGH 3
DFHI 3
DFGI 3
THESE ARE 15 WAYS BY WHICH TOTAL NUMBER OF WAYS BY WHICH COMMITEE CAN BE FOUND COMES OUT TO BE 37(ANS)

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puneet khandelwal  (10 years ago)

1.I-4MEMBERS;II-3MEMBERS;III-2MEMBERS
I CANT HAVE ABC SO NO OF WAYS OF CHOOSING 4 MEMBERS OUT OF 6 MEMBERS(DEFGHI)IS6C4=15;
NOW 15 VALID WAYS ARE
I II III NO OF WAYS OF DOING AS PER QUES
DEFG 3
FGHI 1
EGHI 1
DGHI 1
GDEF 3
HDEF 3
IDEF 3
DEGH 3
DEHI 3
EFGH 1
EFGI 3
EFHI 3
DFGH 3
DFHI 3
DFGI 3
THESE ARE 15 WAYS BY WHICH TOTAL NUMBER OF WAYS BY WHICH COMMITEE CAN BE FOUND COMES OUT TO BE 37(ANS)
2.THERE ARE 15 WAYS BY WHICH III CAN HAVE MEMBERS WHICH ARE AB,AC,BC,DE,DF,EF,FA,FB,FC,DA,DB,DC,EA,EB,EC

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