Permutations & Combinations - Questions & Discussions

MBA CAT 2022
188
please help me with the following question

In how many ways 5 CA students and 6 law students can be arranged together so that no two CA students are side by side?
(a) 7!6!/2! (b) 6!*6! (c) 5!6! (d) 11P5.

i am getting 6!*7P5, as the answer. please do post your appproach


ur answer is correct in my opinion. I feel all options are wrong.
189
how many 4 digit number divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 and 6.
a. 220
b. 588
c. 432
d. 216

Even for this prob. ,i feel options given are not correct.
190
swaminathan21 Says
Even for this prob. ,i feel options given are not correct.



i am getting 390 as ans.(for 4 digit number wala ques) , whats ur ans.:lookround:
191
please help me with the following question

in how many ways 5 ca students and 6 law students can be arranged together so that no two ca students are side by side?
(a) 7!6!/2! (b) 6!*6! (c) 5!6! (d) 11p5.

I am getting 6!*7p5, as the answer. Please do post your appproach

6!*7p2= 6!*7!/2!= 7!*6!/2!
192
ajinder89 Says
i am getting 390 as ans.(for 4 digit number wala ques) , whats ur ans.:lookround:

could u please explain ur method....
193

588
6 ways for 4th digit
7 for second and third
2 for first

194
ajinder89 Says
i am getting 390 as ans.(for 4 digit number wala ques) , whats ur ans.:lookround:

Me too getting 390

case1:
when the last digit is 0:

here the last place can be filled in 1 way.
first can be filled in 7 ways.
2nd in 6 ways.
3rd in 5 ways.
so total ways=7*6*5*1=210.

Case2:
when last digit is 5:
here too last place can be filled in 1 way.
first place can be filled in 6 ways( as 0 cant come here)
2nd place can be filled in 6 ways.(0 may be accomodated here).
3rd place in remaining 5 ways.
so total ways in this case=6*6*5*1=180.

total ways=case1+case2=210+180=390
195
Me too getting 390

case1:
when the last digit is 0:

here the last place can be filled in 1 way.
first can be filled in 7 ways.
2nd in 6 ways.
3rd in 5 ways.
so total ways=7*6*5*1=210.

Case2:
when last digit is 5:
here too last place can be filled in 1 way.
first place can be filled in 6 ways( as 0 cant come here)
2nd place can be filled in 6 ways.(0 may be accomodated here).
3rd place in remaining 5 ways.
so total ways in this case=6*6*5*1=180.

total ways=case1+case2=210+180=390

your method is wrong,
it will contain some redundant numbers

like think u have numbers 5,6,6 and u are req. to make 3 digits numbers from this divisible by 5

so acc. to ur method it will be

first place can be filled by 2 ways(6,6)
second place can be filled by 1 way(6)

so total ways 2*1*1= 2

BUT THE ONLY NUMBER possible is 665

I think u must have got my point.
rethink abt it again

according to me the answer must be 249



whts the OA??
196
how many 4 digit number divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 and 6.
a. 220
b. 588
c. 432
d. 216

Numbers that ends with 0:-

i) With one or no six
6*5*4 = 120 numbers

ii) With two six
Remaining one digit can be chosen in 5 ways
=> 3*5 = 15 such numbers

Numbers that ends with 5:-

i) With one or no six:-
5*5*4 = 100 numbers

ii) With two six
a) With 0
6605 and 6065 - 2 numbers
b) Without 0
Remaining one digit can be chosen in 4 ways
=> 3*4 = 12 numbers

Total = 120 + 15 + 100 + 2 + 12 = 249
197
Numbers that ends with 0:-

i) With one or no six
6*5*4 = 120 numbers

ii) With two six
Remaining one digit can be chosen in 5 ways
=> 3*5 = 15 such numbers

Numbers that ends with 5:-

i) With one or no six:-
5*5*4 = 100 numbers

ii) With two six
a) With 0
6605 and 6065 - 2 numbers
b) Without 0
Remaining one digit can be chosen in 4 ways
=> 3*4 = 12 numbers

Total = 120 + 15 + 100 + 2 + 12 = 249

yeah this is right....
198
An easier one : -

In how many ways 7^311 could be represented as the product of three natural numbers ?

Sayonara

my take for this is 48828

whts the OA???
199
my take for this is 48828

whts the OA???


a*b*c = 7^311
a = 7^x
b = 7^y
c = 7^z

=> x + y + z = 311

Total number of ways = C(313, 311) = 48828

Had we have been asked to find the number of ordered pairs 48828 will be the answer. But we have to find the number of ways in which 7^311 can be written as product of three natural numbers that means 7*7^100*7^210 and 7^210*7^100*7 are same.

Cases when a, b, c all are different are counted 6 times, while when two of them are equal are counted thrice.

Number of cases when two of the three numbers are same = 156

=> Required answer = (48828 - 156*3)/6 + 156 = 8216
200
One more interesting question on probability:
Out of 21 tickets marked with numbers from 1 to 21 ,three are drawn at random, find the probability that the three numbers on them are in A.P
ans:- 10/133

i got 11/133

whts the OA and also tell the method... plz
201
a*b*c = 7^311
a = 7^x
b = 7^y
c = 7^z

=> x + y + z = 311

Total number of ways = C(313, 311) = 48828

Had we have been asked to find the number of ordered pairs 48828 will be the answer. But we have to find the number of ways in which 7^311 can be written as product of three natural numbers that means 7*7^100*7^210 and 7^210*7^100*7 are same.

Cases when a, b, c all are different are counted 6 times, while when two of them are equal are counted thrice.

Number of cases when two of the three numbers are same = 157

=> Required answer = (48828 - 157*3)/6 + 157 = 8216

ohhh yes i made a mistake thanks for ur correction but please explain me how u got that 157...... can`t fetch that.... forgive me
202
i got 11/133

whts the OA and also tell the method... plz


No of AP's with diff 1 - 19
No of AP's with diff 2 - 17
No of AP's with diff 3 - 15
.
.
No of AP's with diff 10 - 1
Total = (1 + 3 + 5 + .. + 19)/C(21, 3) = 10/133

Alternatively:-
When three terms are in AP, middle term is average of the other two terms. That means other two terms are either both odd or both even. So, we just have to choose two even numbers or two odd numbers and we will automatically get the third term

So, P = {C(10, 2) + C(11, 2)}/C(21, 3) = 10/113

Arpitgs Says
ohhh yes i made a mistake thanks for ur correction but please explain me how u got that 157...... can`t fetch that.... forgive me


Sorry its 156 and not 157.

See, if two are equal, then
2x + z = 311
x can vary from 0 to 155, so 156 different values are possible
203
No of AP's with diff 1 - 19
No of AP's with diff 2 - 17
No of AP's with diff 3 - 15
.
.
No of AP's with diff 10 - 1
Total = (1 + 3 + 5 + .. + 19)/C(21, 3) = 10/133

Alternatively:-
When three terms are in AP, middle term is average of the other two terms. That means other two terms are either both odd or both even. So, we just have to choose two even numbers or two odd numbers and we will automatically get the third term

So, P = {C(10, 2) + C(11, 2)}/C(21, 3) = 10/113



Sorry its 156 and not 157.

See, if two are equal, then
2x + z = 311
x can vary from 0 to 155, so 156 different values are possible

thank u so much.....
204
There are 5 blue balls, 2 red, 2 green, 1 black & 1 white. In how many ways can we select 4 out of these 11 balls ?

30
31
35
330
331

:

my take on 31
205
in the first case the answer will be (32c (24c (16c(8c
= (32!)/ (8!)^4
in the second case the parcels are non distinguishable, hence the answer will be [ (32c (24c (16c(8c]/ 4! = (32!)/ {(8!)^4. 4!}

1) c(32,8 )*c(24,8 )*c(16,8 )*4!
2) c(32,8 )*c(24,8 )*c(16,8 )
206
Hi.. is the answer 26.. can't be that str8forward.. or am I missing sumthin.
Lemme know if I'm right
Cheerz

HOW u got 26, it must be 24... post ur method
207
Numbers that ends with 0:-

i) With one or no six
6*5*4 = 120 numbers

ii) With two six
Remaining one digit can be chosen in 5 ways
=> 3*5 = 15 such numbers

Numbers that ends with 5:-

i) With one or no six:-
5*5*4 = 100 numbers

ii) With two six
a) With 0
6605 and 6065 - 2 numbers
b) Without 0
Remaining one digit can be chosen in 4 ways
=> 3*4 = 12 numbers

Total = 120 + 15 + 100 + 2 + 12 = 249



i didnt get the second part, whay calculation of zero is don seperately, kindly explain, thanks