Latest M4maths Aptitude Question SOLUTION: How many combinations are possible if 3 one rupee coin, 4 fifty paise coin and 2 twenty five paise coin are available. It is compulsory to select any one of the coin

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(#M40035100) M4MATHS Numerical Ability question Keep an EYE

How many combinations are possible if 3 one rupee coin, 4 fifty paise coin and 2 twenty five paise coin are available. It is compulsory to select any one of the coin

Asked In M4maths harshal agrawal (10 years ago)

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p.p.singh  (10 years ago)

Total possible combinations = (3+1)*(4+1)*(2+1)=60
But It is compulsory to select any one of the coin.
so 0 one rupee coin, 0 fifty paise coin and 0 twenty five paise coin selection is not allowed.
so desired combinations = 60-1=59

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Udhaya kumar  (10 years ago)

(3c1*4c0*2c0)+(3c0*4c1*2c0)+(3c0*4c0*2c1) = 9................

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RITESH KUMAR  (9 years ago)

we can choose any coin in two way whether we are choosing it or not...i.e.for selecting any coin we have two option
so total combination of selection of zero or more coins=
2+2+2+2+2+2+2+2+2=18
but we have to selct at least one
so ans is 18-1=17

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Divyasri  (9 years ago)

we must take atleast one coin from each
they don't give overally,,how many coins we take
so,we consider all possible combinations
solution is:
[(3c1+3c2+3c3)*4c1*2c1]+[3c1*(4c1+4c2+4c3+4c4)*2c1]+[3c1*4c1*(2c1+2c2)]+
[(3c1+3c2+3c3)*4c2*2c1]+[3c2*(4c1+4c2+4c3+4c4)*2c1]+[3c2*4c1*(2c1+2c2)]+
[(3c1+3c2+3c3)*4c3*2c1]+[3c3*(4c1+4c2+4c3+4c4)*2c1]+[3c3*4c1*(2c1+2c2)]+
...........................................................
like that we will take all possible situations
but, it is very difficult to calculate this..
so,any one please tell me the small logic to this problem

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Pratima mishra  (6 years ago)

total coins=9
so combination =9!=362880

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