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

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.
9 years agoHelpfull: Yes(15) No(0)
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.
9 years agoHelpfull: Yes(3) No(0)
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
9 years agoHelpfull: Yes(2) No(0)
Ans must be option A.
13c1*13c1*13c1*13c1*48c2
9 years agoHelpfull: Yes(2) No(3)
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.
9 years agoHelpfull: Yes(0) No(1)

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)