Permutations & Combinations - Questions & Discussions

MBA CAT 2022
852
@farziPandit said:
In 2nd case repetitions are taking place
eg. You choose B1 & G1 first then selecting B4,G2 OK
Then in another case you choose B1 & G2 first then selecting B4 & G1.......
853

thnx dude!!

854

Kindly help me with this one:


In how many ways we can select four cards of different suits & different denominations from a deck of 52 cards??

Confusion is between : 52 X 36 X 22 X 10
and 13 X 12 X 11 X 10

Which one is correct ???

@LeoN88 Can Anyone of you help me out with this one??
855
@farziPandit said:
Kindly help me with this one:In how many ways we can select four cards of different suits & different denominations from a deck of 52 cards??Confusion is between : 52 X 36 X 22 X 10 and 13 X 12 X 11 X 10 Which one is correct ??? @LeoN88Can Anyone of you help me out with this one??
13x12x11x10

start posting on quant thread if you want an instant reply.
856
@gs4890 said:
13x12x11x10start posting on quant thread if you want an instant reply.
bhai 52*36*22*10 hoga na. to choose a suit also.
857
@maddy2807 said:
bhai 52*36*22*10 hoga na. to choose a suit also.
13 cards in each suit.

Now, if you select any one of the 13 cards of say spades ; u'll be left with only 12 options to choose from the next set of suit & so on.........

Arun sharma ka qstn hain bhai ; kiya hua hain boht baar.

baaki dekh lo ....letz see what @krum & @rkshtsurana says
858
@farziPandit said:
Kindly help me with this one:In how many ways we can select four cards of different suits & different denominations from a deck of 52 cards??Confusion is between : 52 X 36 X 22 X 10 and 13 X 12 X 11 X 10 Which one is correct ??? @LeoN88Can Anyone of you help me out with this one??
bhai 4 suits bole to 13,13,13,13

choose 1 from any1 - 13c1
ab baki sab se 1 option kam, i.e 12c1, similarly

13c1*12c1*11c1*10c1

859
@krum said:
bhai 4 suits bole to 13,13,13,13choose 1 from any1 - 13c1ab baki sab se 1 option kam, i.e 12c1, similarly13c1*12c1*11c1*10c1
bhai my doubt is shud me choose suits also?
860
@farziPandit said:
Kindly help me with this one:In how many ways we can select four cards of different suits & different denominations from a deck of 52 cards??Confusion is between : 52 X 36 X 22 X 10 and 13 X 12 X 11 X 10 Which one is correct ??? @LeoN88Can Anyone of you help me out with this one??
1st select any card = 52C1
now we have left with 52 - 13 - 3 = 36 cards
2nd card = 36C1
in same way other 2 in 22C1 and 10C1

total ways = 52C1*36C1*22C1*10C1

but here order doesn't matter so , we have to divide it by 4!

required answer will be = 52C1*36C1*22C1*10C1/4!
861
@jain4444 said:
1st select any card = 52C1 now we have left with 52 - 13 - 3 = 36 cards 2nd card = 36C1 in same way other 2 in 22C1 and 10C1 total ways = 52C1*36C1*22C1*10C1 but here order doesn't matter so , we have to divide it by 4! required answer will be = 52C1*36C1*22C1*10C1/4!
Are you sure?
Check this:

You can select 4 different numbers in 13C4 ways
Now you have to assign 4 different suits to these 4 numbers: which you can do in 4! ways

Total no. of ways: 13C4 * 4!
862
@smiyc said:
Are you sure?Check this:You can select 4 different numbers in 13C4 waysNow you have to assign 4 different suits to these 4 numbers: which you can do in 4! waysTotal no. of ways: 13C4 * 4!

Both answers are same buddy :)
863
@jain4444 said:
Both answers are same buddy
Are they ?? Then might be I was calculating your answer incorrectly ...
864
@jain4444 said:
Both answers are same buddy

Agreed. There are two ways to reach the solution-

i) C(52,1)*C(36,1)*C(22,1)*C(10,1)/4!
ii) C(13,1)*C(12,1)*C(11,1)*C(10,1)
865

8 balls of different colours are arranged in a row.

Condition- red and white may not come together.
866
@macc6 said:
8 balls of different colours are arranged in a row.Condition- red and white may not come together.
_ RED _ WHITE _

g1 + g2 + g3 = 6 (where g2 > 0)
=> number of solutions = 7C2 = 21

and in 2! ways red and white arrange themselves

required answer = 21*6!*2!
867
@jain4444 Can you tell me what is g1,g2,g3??
868

Kindly help

The no of ordered triplets of +ve inegers which are solution of the equation~ x+y+z=100
869
@macc6 take all cases and subtract those in which red and white come together.
So, total cases = 8!
Cases in which red and white come together = 7!*2!
So, required answer = 8! - 7!*2!
870
@macc6
 x + y + z = 100;
0 00
x , y, and z can only be from 1 to 98
Because if the other two nos, are both 1, the value of the third is 98.

if x = 1, y could only be 1 to 98 (98 choices)
if x = 2, y could only be 1 to 97 (97 choices)
......
If x = 98, c could only be 1 (1 choice)
Once x and y are selected, z could only be 100 - (x+y)
Se we only count the various x, y combinations

Total number of ordered triplets
= 98 + 97 + .... + 2 + 1
= 98(98+1)/2 = 4851

(summation of integers from 1 to n = n(n+1)/2
n = 98
871
@mayurdhingra thnxx a lot...