Permutations & Combinations - Questions & Discussions

MBA CAT 2022
550
please answer dis-
how many numbers which are divisible by 5 can be formed using the digits 0 to 6 if no digits occur more than once in any number?


Since numbers those are divisible by 5 end with 0 and 5 we will take two scenarios here:
1.Numbers ending with 0.

if unit digit contains 0 then all other digits an be arranged in 6! ways.

OR
2. Numbers ending with 5
If unit digit consist of 5 then _ _ _ _ _ _ 5 then the first digit from left can be filled with 5 numbers except 0 then otherdigits can be filled with 5! ways.

So it will be 1320...plz correct me if iam missing something..
551
Since numbers those are divisible by 5 end with 0 and 5 we will take two scenarios here:
1.Numbers ending with 0.

if unit digit contains 0 then all other digits an be arranged in 6! ways.

OR
2. Numbers ending with 5
If unit digit consist of 5 then _ _ _ _ _ _ 5 then the first digit from left can be filled with 5 numbers except 0 then otherdigits can be filled with 5! ways.

So it will be 1320...plz correct me if iam missing something..

if 4-digits number are to be formed???
552
Surbhih Says
if 4-digits number are to be formed???

Proceed the same as explained above

Case I : Last digit 0

So 6*5*4 i.e. 120

Case II: Last digit 0

So 5*5*4 i.e. 100

Therefore total required numbers is 220

correct me if I am wrong
553
Proceed the same as explained above

Case I : Last digit 0

So 6*5*4 i.e. 120

Case II: Last digit 0

So 5*5*4 i.e. 100

Therefore total required numbers is 220

correct me if I am wrong

u hav written 2nd case also of 0 and dat was my only doubt in dis question?
554
please answer dis-
how many numbers which are divisible by 5 can be formed using the digits 0 to 6 if no digits occur more than once in any number?


last digit has to be either 0 or 5

when last digit is 0 :-

_ _ _ _ 0

now we can fill 1st blank spaces with 6 numbers i.e 1 to 6
2nd with 5
3rd with 4
4th with 3

so , 6*5*4*3 = 360

when last digit is 5 :-

_ _ _ _ 5

now 1st spaces can be filled with 5 numbers i.e. 1,2,3,4,6
2nd with 5 numbers because we will also include 0 here

so , 5*5*4*3 = 300

total = 360+300
555
Surbhih Says
u hav written 2nd case also of 0 and dat was my only doubt in dis question?

OOPs sorry.. typo

the second case is for last digit 5

and the calculation is correct
556
Q) There are 6 points on a circle. five events occur at 5 of these points ,one at each point. in how many ways can they occur such that no two successive events occur at adjacent points?
557

total arrangements will be 6P5.....now we will take scenario where we will consider two point as one...so it will be 5C5 x 2!...
Ans- 6P5 - 5C5 x 2!


ps: correct me if iam making some mistake in this question..

558
total arrangements will be 6P5.....now we will take scenario where we will consider two point as one...so it will be 5C5 x 2!...
Ans- 6P5 - 5C5 x 2!


ps: correct me if iam making some mistake in this question..


this is not d answer
559
Surbhih Says
Q) There are 6 points on a circle. five events occur at 5 of these points ,one at each point. in how many ways can they occur such that no two successive events occur at adjacent points?


My take is 14*5! (Assuming distinct events are to be premuted)

Please check this post:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a-890.html#post3222377
560

I really didnt get the answer ThinkAce .
Could you elaborate in detail .

561
In a bag there are 3 green coloured balls, 5 red coloured balls, 2 blue coloured balls, 6 black coloured balls.

1)If 5 balls are selected at random from the bag,wht is the probability that the selection contains 2 blue balls and 3 red balls?
a)7C5/16C5 b)20/16C5 c)10/16C5 d)30/16C5 e)none

2)If 4 balls are selected at random from the bag then wht is the probability that the selection contains either red balls or black balls?
a)10C4/16C4 b)(5C4*6C4)/16C4 c)(5C4 +6C4)/16C4 d)11C4/16C4 e)none

i got ma ans as:
1) b
2) e

plzzzzzzzzzzzzz solve n c wht ans u people get!!!



1) Total balls 16 out of it we have to select 2 blue and 3 red balls
so selection = 2c2 * 5c3 = 10
Total selection =16c5
so ans is 10/16c5


2) You must select 4 balls either of red or black colors
so selection = ( 5c4 + 6c4 )
Total selection = 16c4
Ans is ( 5c4 + 6c4 )/ 16c4
562
1) Total balls 16 out of it we have to select 2 blue and 3 red balls
so selection = 2c2 * 5c3 = 10
Total selection =16c5
so ans is 10/16c5


2) You must select 4 balls either of red or black colors
so selection = ( 5c4 + 6c4 )
Total selection = 16c4
Ans is ( 5c4 + 6c4 )/ 16c4



I agree

its B and C
563

Hi I have a doubt...can anyone ans this...
There are 10 questions in a class test.In how many ways can a student solve one or more questions.
options a) 1023 b)1024 c)1025 d)1000 e)None of these

564
Hi I have a doubt...can anyone ans this...
There are 10 questions in a class test.In how many ways can a student solve one or more questions.
options a) 1023 b)1024 c)1025 d)1000 e)None of these


There are 2 ways to attempt each question - to do it or not to do it.

He can solve 10 questions in 2^10 ways. However, out of this remove the case where he does not attempt any.

Hence, possible cases are 2^10 - 1 = 1024-1 - 1023
565

Hi..Thanks for the reply..I have one more query.
In a race the probability that A will win the race is 1/2, the probability that B wins the race is 1/4 and the probability that C will win is 1/5. Assuming dead heat impossible, what is the chance that one of them will win the race?
a)1/20 b)1/40 c)1/11 d)19/20 e)indeterminate

566
hi..thanks for the reply..i have one more query.
In a race the probability that a will win the race is 1/2, the probability that b wins the race is 1/4 and the probability that c will win is 1/5. Assuming dead heat impossible, what is the chance that one of them will win the race?
A)1/20 b)1/40 c)1/11 d)19/20 e)indeterminate


= (1/2+1/4+1/5)

=(10+5+4/20)

=19/20
567
Hi..Thanks for the reply..I have one more query.
In a race the probability that A will win the race is 1/2, the probability that B wins the race is 1/4 and the probability that C will win is 1/5. Assuming dead heat impossible, what is the chance that one of them will win the race?
a)1/20 b)1/40 c)1/11 d)19/20 e)indeterminate


As the outcomes are not dependent and dead heat (meaning two or all the three of them finish the race at the same time) is not possible.
Hence, probability = 1/2 + 1/4 + 1/5 => (10+5+4)/20 = 19/20
568

Hi,
My query is :

10 persons seated in a row find d ways in which 4 persons can be selected such dat no 2 of them are sitting nex to each oder?


Thanks

569
Hi,
My query is :

10 persons seated in a row find d ways in which 4 persons can be selected such dat no 2 of them are sitting nex to each oder?


Thanks

what is the OA, is it 1920