Latest Book Aptitude Question SOLUTION: From a pack of 52 playing cards, 4 cards are removed at random. In how many ways can the 1st place and 3rd place cards be drawn out such that both are black ?

There are 26 black cards to choose from on the first draw.
Now the second card can either be black or red.

If black, then there are 25 to choose from, and then for
the third card there will then be 24 black cards to choose
from.

If red, then there are 26 reds to choose from, and then
for the third card there are then 25 black cards to
choose from.

In either case, there are 49 cards left to from on the
4th draw.

So the number of ways we can do this is

26*(25*24 + 26*25)*49 = 1,592,500,

so option 4 is correct.

Among the four cards , the first and the third card must be a black one.
First card can be chosen from 26C1 .... Combination
Third card can be chosen 25C1..... (1 Black Card eliminated)

Now the second and the fourth cards
Second card can be chosen from 50C1....(2 Cards eliminated)
Forth card can be chosen from 49C1.... (3 Cards eliminated)

So, the Results would be 26*25*50*49 = 1592500

Half of the deck is black, so there are 52/2 = 26 ways to choose the first black card. The next bit is a little tricky.

You have drawn one card so there are 51 left. However, how many can be black depends on what you drew the second time (we are already assuming the first was black).

If the second card you draw is red, then there are 26 red cards to choose from: 26*26 = 26² choices for the 1st two AND there are still 25 (26 - 1) choices for the third card to be black.

If the second card is black, however, there are only 25 choices, so the number of ways to choose the first two as black is 26*25 (instead of 26²). Additionally, there are now only 26 - 2 = 24 black cards remaining so the number of ways to choose this third black card is actually 24, not 25 like above.

Finally, in both cases, the last card is any of the 52 - 3 = 49 cards left (regardless of whether the 2nd card was black or red).

The two are mutually exclusive: you cannot both draw a red 2nd AND draw a black second (you could draw a diamond second AND draw a red card second, for example). When things are mutually exclusive and you simply add them:

So you get:

26² * 25 * 49 + 26*25*24*49 = 49*26*25 * (26 + 24) = 1,592,500

----26-----50-----25------49----

THAT CAN BE ARRANGE IN DIFFERT WAYS WE JUST NEED TO MULTIPLY THAT ALL NUMNBER
26 IS 1ST TIME(1ST POSITION) 50 IS 2ND POSITION SO ON .............