Permutations & Combinations - Questions & Discussions

MBA CAT 2022
612
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613

Could anyone please answer this?? What is the number of ways to arrange 4 red and 6 green balls in a circle? Given that balls of same color are considered same. If your answer is 21. Try the same procedure for 2 red balls and 2 green balls!!

614
PAUL KINGSLEY Says
Could anyone please answer this?? What is the number of ways to arrange 4 red and 6 green balls in a circle? Given that balls of same color are considered same. If your answer is 21. Try the same procedure for 2 red balls and 2 green balls!!


As,per me it should be 9! / 3!6! + 9!/ 4!5!
615

Pls som1 reply....

in how many ways can the letters of the word COUPLE be arranged so as no vowels are 2gather....
(extremely amateurish qstn...just have an iota of doubt abt the answers provided... 😛 )

616
Pls som1 reply....

in how many ways can the letters of the word COUPLE be arranged so as no vowels are 2gather....
(extremely amateurish qstn...just have an iota of doubt abt the answers provided... 😛 )


Can be done by many method.

_O_U_E_ (OUE themselves can be arranged in 3! manner)
a+b'+c'+d = 1
so 4!/3!

Now these remaining 3 can be arranged in 3! manner
so

3!*4! My take.
617

but...i might b wrong...
isnt it suosd to b...suppose..
O_U_E -->3!
for the 2 consonents in the middle... 3C2
and again for the end 2 positns..only one consonent...thus.. 2C1

3! x 3C2 x 2

618

If a fair coin is tossed 114 times what is the probability that atleast 72 outcomes will be tail?

619
Can be done by many method.

_O_U_E_ (OUE themselves can be arranged in 3! manner)
a+b'+c'+d = 1
so 4!/3!

Now these remaining 3 can be arranged in 3! manner
so

3!*4! My take.


but...i might b wrong...
isnt it suosd to b...suppose..
O_U_E -->3!
for the 2 consonents in the middle... 3C2
and again for the end 2 positns..only one consonent...thus.. 2C1

3! x 3C2 x 2
620
Pls som1 reply....

in how many ways can the letters of the word COUPLE be arranged so as no vowels are 2gather....
(extremely amateurish qstn...just have an iota of doubt abt the answers provided... 😛 )


total arrangements = 6!
arrangement in which vowels are together = 4! *3!
required arngmnt = 6! - 4!*3!
= 576
621
PAUL KINGSLEY Says
Could anyone please answer this?? What is the number of ways to arrange 4 red and 6 green balls in a circle? Given that balls of same color are considered same. If your answer is 21. Try the same procedure for 2 red balls and 2 green balls!!

for the first one it is coming 21
(10-1)! / 4!*6!
but for the next part--a fraction
622
total arrangements = 6!
arrangement in which vowels are together = 4! *3!
required arngmnt = 6! - 4!*3!
= 576



Hey,
It is saying that no vowels are together.
That means, no two vowels and no three vowels together.

We have to consider the cases where all the three vowels ( O,U,E) are seperated from each other.

So, it will be 3! * 3! * 2. = 72 ways..

3! => Placing them at odd/even places.
3! => Arranging the remaning three consonants.
2 => Rearranging the odd and even places..
623
PAUL KINGSLEY Says
Could anyone please answer this?? What is the number of ways to arrange 4 red and 6 green balls in a circle? Given that balls of same color are considered same. If your answer is 21. Try the same procedure for 2 red balls and 2 green balls!!


is the answer for 2 red and 2 green is 2..
624

A mother distributes 5 different apples among 8 children.
How many ways can this be done if there is no restriction on the number of apples a child can recieve?

is it 8^5 or 5^8.I am confused. please help

625
A mother distributes 5 different apples among 8 children.
How many ways can this be done if there is no restriction on the number of apples a child can recieve?

is it 8^5 or 5^8.I am confused. please help



Hey,
In these type of question, the elements that you are placing matters the most.

For e.g i say in how many ways can you place three letters in four boxes.
Now we need to place 3 letters .

SO each letter has got four ways( Remember letters will go into the boxes not vice-versa).
Therefore we have 4*4*4 => 4^3.

Similarly in your question above, first apple can be given to 8 children, 2nd to 8 and so on..

Therefore, 8*8*8*8*8 = 8^5.

Hope this helps. Thanks. 😃
626

Thanks Jai for the above post.It helps

627
PAUL KINGSLEY Says
Could anyone please answer this?? What is the number of ways to arrange 4 red and 6 green balls in a circle? Given that balls of same color are considered same. If your answer is 21. Try the same procedure for 2 red balls and 2 green balls!!

for the first one it is coming 21..
for 2 red and 2 green..is ans 2?
628

Guys I have a dbout in one basic logic if there are 4 items 2 of one type and two of another type then is the number of ways of aranging them is 4!/(2!) or it is
4!/(2! * 2!)

629
Guys I have a dbout in one basic logic if there are 4 items 2 of one type and two of another type then is the number of ways of aranging them is 4!/(2!) or it is
4!/(2! * 2!)


it will be 4!/(2! * 2!)
630

7) In how many ways can we choose a black and a white square on a chess board if the two squares do not belong to the same row or column?

is the answer 64*24 or 32*24 ....am confused

631
7) In how many ways can we choose a black and a white square on a chess board if the two squares do not belong to the same row or column?

is the answer 64*24 or 32*24 ....am confused



32*24

out of 32 balck squares u will choose one black sqaure in 32C1 ways.

Now the white squares in same row and column will not be selected so out of 32 white squares we will take out 8 white squares which are in same row or column as is balck square. so out of remaining 24 we will select 1 white square in 24C1 ways

32*24

The above will be the case when it is given that only black square has to be selected first but when it is given any colour square out of black or white can be choosen first than it will be

64*24