Latest Elitmus Aptitude Question SOLUTION: In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present? a. (13^4) x 48 x 47 b. (13^4) x 27 x 47 c. 48C6 d. 13^4 e. 13^4(6^3 + 88)

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(#M40117694) ELITMUS Numerical Ability question Probability Keep an EYE

In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

a. (13^4) x 48 x 47

b. (13^4) x 27 x 47

c. 48C6

d. 13^4

e. 13^4(6^3 + 88)

Asked In Elitmus Dolly Singh (9 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

Surit Datta  (9 years ago)

There are 13 cards in each suit and 52 cards and 4 suits in a deck.

1 -card (representative of 1st suit:) can be chosen in 13 ways
2- card (representative of 2nd suit) again in 13 ways
3- card (representative of 3rd suit) again in 13 ways
4- card (representative of 4th suit) also in 13 ways

and there remain 48 cards in a deck and Remaining 2 cards can be chosen in 48C2 ways.

So 13*13*13*13* 48C2 or 13^4* 48C2.

Is this solution Helpfull? Yes (34) | No (4)  

shiv  (9 years ago)

Here in a deck there are 52 cards.
in every suits here are 13 cards.
we have to draw the card one by one one without replacement.
so for the 1st card of 1st suit we have 13 ways.
so for the 1st card of 2nd suit we have 13 ways.
so for the 1st card of 3rd suit we have 13 ways.
so for the 1st card of 4th suit we have 13 ways.
now we have 48 cards.
we have to draw two cards one by one
no of ways=48*47
total no of ways=13*13*13*13*48*47.
OPTION A is the correct ans.

Is this solution Helpfull? Yes (15) | No   

priti  (9 years ago)

from 13 cards we have to choose 1,like that 4 times,then from remaining cards we have to take 1 means 48 ways then 47 remaining ways to choose one more.

Is this solution Helpfull? Yes (3) | No   

Rajmohan  (9 years ago)

ans is (13^4) x 48c2
=(13^4) x 47 x 24
coz it is just ways of selection
for permutation = (13^4) x (48c2).!2 = (13^4) x 47 x 48

Is this solution Helpfull? Yes (2) | No   

ASHISH MUKHERJEE  (9 years ago)

Ans must be option A.
13c1*13c1*13c1*13c1*48c2

Is this solution Helpfull? Yes (2) | No (3)  

Ayush Gupta  (9 years ago)

There are 13 cards in each suit and 52 cards and 4 suits in a deck.

1 -card (representative of 1st suit:) can be chosen in 13 ways
2- card (representative of 2nd suit) again in 13 ways
3- card (representative of 3rd suit) again in 13 ways
4- card (representative of 4th suit) also in 13 ways

and there remain 48 cards in a deck and Remaining 2 cards can be chosen in 48C2 ways.

So 13*13*13*13* 48C2 or 13^4* 48C2.

Thus 6 cards represent all suits.

Is this solution Helpfull? Yes | No (1)  

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