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.
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
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.
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
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
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
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
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
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
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!}