@khareaditya25
Every chocolate can be distributed in 3 ways so the total no. of ways should be 3*3*3*3*3*3=3^6=729
I hope this is correct.
@priyagoyal2711 said:It will help, if you guys can tell me what methods you used to solve then, and then it can be figured out.
Let a,b and c be the no of chocolates given to 3 person with a condition that a,b,c=0 and a=5-a' , b=5-b' and c=5-c'.Now to get the answer we have to find out the solution for
a+b+c=10
replacing the values of a.b and c we get
15-(a'+b'+c')=10
-> a'+b'+c'=5
which will give the solution as 5+3-1C3-1=21 ways
I belive the solution given by 2ndApr2011 is more explanatory than mine, but this was the way i got mine. Hope the ans is correct atleast (if not the method
)@saurabh2996 and what changes will be there if each boy has to get at least 1 ?
To find the solution with this condition, we need to substract the cases where any one of them would have got 0 chocolates. Below are the four condition where atleast one of them got 0 chocolates:
A B C No. of ways
0 0 6 6C6*3=3
0 1 5 6C1*3!=36
0 2 4 6C2*3!=90
0 3 3 6C3*3=60
--------------------------------------------------------
Total No of ways= 189
So total no of ways in which every person get atleast 1 chocolate is 729-189=540
@khareaditya25: Please let me know if the answer is correct.
PUYs......please share your solution if anyone has an alternate solution to this problem.
Cheers!!!!
A B C No. of ways
0 0 6 6C6*3=3
0 1 5 6C1*3!=36
0 2 4 6C2*3!=90
0 3 3 6C3*3=60
--------------------------------------------------------
Total No of ways= 189
So total no of ways in which every person get atleast 1 chocolate is 729-189=540
@khareaditya25: Please let me know if the answer is correct.
PUYs......please share your solution if anyone has an alternate solution to this problem.
Cheers!!!!
@saurabh2996
For the password problem
Case 1a - 2 digits 3 vowels and begins with digit ends with vowel
First slot can be filled in 3c1 ways = 3
5th slot can be filled in 5c1 ways = 5
For the middle three,
Select two vowels from the remaining four = 4c2 = 6
Select one digit from the remaining two = 2c1 = 2
The middle three can be arranged in 3! ways = 6
So, total ways = 3x5x6x2x6 = 1080
For the password problem
Case 1a - 2 digits 3 vowels and begins with digit ends with vowel
First slot can be filled in 3c1 ways = 3
5th slot can be filled in 5c1 ways = 5
For the middle three,
Select two vowels from the remaining four = 4c2 = 6
Select one digit from the remaining two = 2c1 = 2
The middle three can be arranged in 3! ways = 6
So, total ways = 3x5x6x2x6 = 1080
Case 1b - 2 digits 3 vowels and begins with Vowel ends with digit
Everything same as done in the above case just that the first slot can be filled in 5 ways and last slot in 3 ways. Everything same for the middle three.
No. of ways = 5x3x6x2x6 = 1080
First slot, 3c1 = 3
Last slot, 5c1 = 5
For middle three,
Select two digits from remaining two in 2c2 ways = 1
Select one vowel from remaining four in 4c1 ways = 4
Middle three can be arranged in 3! ways = 6
Total ways = 3x5x1x4x6 = 360
Everything same as done in the above case just that the first slot can be filled in 5 ways and last slot in 3 ways. Everything same for the middle three.
No. of ways = 5x3x6x2x6 = 1080
Case 2a - 3 digits 2 vowels and begins with digit ends with vowel
Last slot, 5c1 = 5
For middle three,
Select two digits from remaining two in 2c2 ways = 1
Select one vowel from remaining four in 4c1 ways = 4
Middle three can be arranged in 3! ways = 6
Total ways = 3x5x1x4x6 = 360
Case 2b - 3 digits 2 vowels and begins with Vowel ends with digit
Everything same as Case 2a, just that first slot can be done in 5c1 and last can be done in 3c1. Everything is same for the middle three.
Total ways = 5x3x1x4x6 = 360
Total = 1080+1080+360+360 = 2880
Everything same as Case 2a, just that first slot can be done in 5c1 and last can be done in 3c1. Everything is same for the middle three.
Total ways = 5x3x1x4x6 = 360
Total = 1080+1080+360+360 = 2880
guys .. i am having a problem here...
A buys 6 distinct choclates to be distributed among P. Q, R. Each has to get 2 choclates on a condition that P does not like Melody and Q does not like Eclairs. In how many ways can they be distributed?
@pusghatole said:guys .. i am having a problem here...A buys 6 distinct choclates to be distributed among P. Q, R. Each has to get 2 choclates on a condition that P does not like Melody and Q does not like Eclairs. In how many ways can they be distributed?
Is the ans 78?? If it is correct I'll explain it,else it is -1 for me 
@prati086 said:How many words cn be formed with word LUCKNOW which havei)L always occuring before Uii)L always occuring before U and U always occuring before WGuys/Gals expecting detailed approach fr ths question in Lucid Manner
Are the ans
i. 1920
ii. 860
??
i. 1920
ii. 860
??
@pusghatole said:@saurabh2996 man... the correct answer is 42.... i am still searching the method...on mars...
this needs a deeper thought!!!@prati086 said:How many words cn be formed with word LUCKNOW which havei)L always occuring before Uii)L always occuring before U and U always occuring before WGuys/Gals expecting detailed approach fr ths question in Lucid Manner
i) 7!/2 = 2520
ii) 7!/4 = 1260 ??
@pusghatole said:@saurabh2996 man... the correct answer is 42.... i am still searching the method...on mars...
I believe the solution has to be this way:
The total number of ways in which 6 Different chocolates can be distributed to 3 persons (Namely P, Q and R) is 6C2*4C2*2C2=90 Ways
We know that P does not like Melody and R does not like Eclairs so we need to remove those ways from the total no of ways where P and R get the chocolates which they don't like:
CASEI:
When P has already received Melody which he doesn't like. So we are left with 5 Chocolates which can be distributed to P,Q and R resp in 5C1*4C2*2C2=30 Ways
CASE II:
The total number of ways in which 6 Different chocolates can be distributed to 3 persons (Namely P, Q and R) is 6C2*4C2*2C2=90 Ways
We know that P does not like Melody and R does not like Eclairs so we need to remove those ways from the total no of ways where P and R get the chocolates which they don't like:
CASEI:
When P has already received Melody which he doesn't like. So we are left with 5 Chocolates which can be distributed to P,Q and R resp in 5C1*4C2*2C2=30 Ways
CASE II:
When R has already received Eclairs which he doesn't like. So we are left with 5 Chocolates which can be distributed to P,Q and R resp in 5C2*3C2*1C1=30 Ways
So total 60 ways in which either P or R gets a chocolate which they do not like. So we are now left with 30 ways. But both the above cases include the case where P and R both got the chocolates they do not like and hence it was substracted twice.So to get to the answer we need to add the no. of ways when both of them got the chocolates they don't like once.
so the no. of ways in which both of them got the chocolates they do not like is 4C1*3C2*1C1=12 Ways
Hence the ans is 30+12=42 Ways
This is what I could think of. I was applying the same logic before but there just substracted the case where both of them got the chocolates they do not like which was 12 and thats why ans was coming as 78. Jaldbazi ke chakkar me hamesha mara jata hun
So total 60 ways in which either P or R gets a chocolate which they do not like. So we are now left with 30 ways. But both the above cases include the case where P and R both got the chocolates they do not like and hence it was substracted twice.So to get to the answer we need to add the no. of ways when both of them got the chocolates they don't like once.
so the no. of ways in which both of them got the chocolates they do not like is 4C1*3C2*1C1=12 Ways
Hence the ans is 30+12=42 Ways
This is what I could think of. I was applying the same logic before but there just substracted the case where both of them got the chocolates they do not like which was 12 and thats why ans was coming as 78. Jaldbazi ke chakkar me hamesha mara jata hun
gr8 work man... another one .......what is the remainder whn 17^17 is divided by 14
@ritika_j Hi: Im new here and havn't gone through the whole thread but for 70-79 there will be 11 7s (77 has 2 7s).. thats probably the difference between your 90 and magis_xl's 91
please help me understand this:
In how many ways can we distribute 6 different chocolates among 3 boys so that each boy receives at least one chocolate?
TIME ANSWER:
6 diff chocolates can be distributed among 3 boys in combination of (114),(123),(222)
ways for (114)=6C4*2*3=90
ways for (123)=6C3*3C2*3!=360
ways for (222)=6C2*4C2=90
no of ways = 540
I solved this way:
no of ways = total number of ways - ways where one of them given 0 chocos - ways where 2 of them given 0 chocos
ANS=3^6 - 3*2^6 - 3 = 534
please tell me what is the flaw in my logic.