sorry chilli bhai.. point out the mistake if you can..
planetabhishek Sayssorry chilli bhai.. point out the mistake if you can..
What you are doing is choosing 5 out of 7 and then permuting them and then placing the remaining 2 in 5*4 ways.
Actually, it should be 5*5 and not 5*4, as the remaining two can go to same pot also.
But its still not correct.
Why??
Say the coins are A, B, C, D, E, F, G and pots are 1, 2, 3, 4, 5
Now, lets consider the following two cases:-
i) Lets first chose A, B, C, D and E and put them in pots 1, 2, 3, 4, 5 respectively.
Now put F and G in 1 and 2 respectively.
So, we have
1 - A, F
2 - B, G
3 - C
4 - D
5 - E
ii) Lets first chose F, B, C, D and E and put them in pots 1, 2, 3, 4, 5 respectively.
Now put A and G in 1 and 2 respectively.
So, we have
1 - A, F
2 - B, G
3 - C
4 - D
5 - E
So, we can see that final result is same, but we have counted it twice may be more than that. that means we can not proceed in this manner.
In how many ways can letters of the word RAINBOW be arranged so that A always comes before I and I always comes before O?
a)840 b)420 c)720 d) none
My approach is -
Arrange 4 letters(R,N,B,W) in 7 places.
7p4
The remaining 3 letters (A,I,O) can be arranged only in 1 way.
So total ways are 7p4 ->840
The ans is correct.I just want to confirm if this approach would work universally for all questions of this sort as the book follows a complicated approach.
Is there a better , easier way?
Thanks in advance!
Your approach is correct but i have an alternative approach which might be helpful too!
The letters of the word 'rainbow' can be arranged in 7p7= 7!=5040 ways. Among themselves , the alphabets , a ,i and o can be arranged in 3!=6 ways. But we need only one form of arrangement of these letters.as such ,the required no of permutaions is 5040/ 6=840.
Thanks.Its always better to have an alternative approach 😃
How many 5 digit no. are possible such that the 2nd digit is greater than the 4th and the 5th digit is greater than the 1st ?
I'm not sure of the OA ... 
How many 5 digit no. are possible such that the 2nd digit is greater than the 4th and the 5th digit is greater than the 1st ?
I'm not sure of the OA ...
abcde is the number, such that b > d and e > a
We can have b > d in C(10, 2) = 45 ways
and e > a in C(9, 2) = 36 ways, as a can not be 0
=> Total such numbers = 45*36*10 = 16200 (here 10 is for c, as c can take any of the 10 digits)
abcde is the number, such that b > d and e > a
We can have b > d in C(10, 2) = 45 ways
and e > a in C(9, 2) = 36 ways, as a can not be 0
=> Total such numbers = 45*36*10 = 16200 (here 10 is for c, as c can take any of the 10 digits)
Can you pls explain why 'a' can't be 0?
dreamsohini SaysCan you pls explain why 'a' can't be 0?
'a' cannot be 0, because, then it would be a 4 - digit number ...
I think its 12.
We can arrange the 4 men in 3! = 6 ways around the circle.
Now, there are just two of arranging the 4 women between the 4 men such that no couple sits together (you can try it manually, it won't take much time)
So, total 6*2 = 12 ways
Answers are out. You got the answer, concept right ! 😃
Your approach is correct but i have an alternative approach which might be helpful too!
The letters of the word 'rainbow' can be arranged in 7p7= 7!=5040 ways. Among themselves , the alphabets , a ,i and o can be arranged in 3!=6 ways. But we need only one form of arrangement of these letters.as such ,the required no of permutaions is 5040/ 6=840.
a different approach..
keep A,I,O in the same order... then we have to arrange the rest of the alphabets in between them or towards their sides. there are 4 such alphabets..
XAXIXOX.. assume X to be the position of the alphabets... assume all of them to be same now..
no of possible ways is 4+4-1C4-1===7C3...
these are the different combinations possibe.. now arrange them by multiplying with 4!..
ans= 7C3*4!== 840..
Hi All,
How many words are there in RAINBOW in which vowels are never together??
Please help.
Kindly explain the answer.
Thanks in advance.
Conditions where the vowels are together= 5!x3!
So OA would be 7!-5!x3!
Hi arunavaray,
Thanx.
But the question says,when vowels are not taken together.
That also leads us to a situation where 2 vowels are also not taken together
except for three.
Won't we count that situatuion too.
The book gives me a differnt method to solve it which i am unable to understand.
If you are aware of the same,then kindly explain the same.
There are three vowels A,I and O.
First of all arrange all the four consonants which can be done in 4! ways.
Now we have 5 places availabe for the three vowels to be filled up, which can be done in 5P3 ways.
Thus total no. of arragnmnts = 4! * 5P3 = 1440
Could you please explain me how this method follows.
Thanx in advance.
Hi All,
How many words are there in RAINBOW in which vowels are never together??
Please help.
Kindly explain the answer.
Thanks in advance.
Conditions where the vowels are together= 5!x3!
So OA would be 7!-5!x3!
Hi arunavaray,
Thanx.
But the question says,when vowels are not taken together.
That also leads us to a situation where 2 vowels are also not taken together
except for three.
Won't we count that situatuion too.
The book gives me a differnt method to solve it which i am unable to understand.
If you are aware of the same,then kindly explain the same.
There are three vowels A,I and O.
First of all arrange all the four consonants which can be done in 4! ways.
Now we have 5 places availabe for the three vowels to be filled up, which can be done in 5P3 ways.
Thus total no. of arragnmnts = 4! * 5P3 = 1440
Could you please explain me how this method follows.
Thanx in advance.
Hi,
Say, C = Consonent, V=Vowel
We have
_ C _ C _ C _ C _
Number of ways to arange consonents = 4!
Now, there are 5 places where we can place vowels (indicated by a dash)
So, number of ways to arrange 3 vowels = 5P3
So, total arrangements = 4! * 5P3 = 1440
P.S.: This problem is same as 'Number of ways to arrange 4 boys and 3 girls with no girls together'.
Hi Puys,
Kindly explain the below.
Sanjog and Rahul took part in a two-day maths contest. At the end both had attempted questions worth 500 marks. Sanjog scored 160 out of 300 attempted on the first day and 140 out of 200 attempted on the second day, so his two-day success ratio was 3/5. Rahul's attempted figures were different from Sanjog's (but with the same two-day total). Rahul had a positive integer marks on each day. For each day Rahul's success ratio was less than Sanjog's. What is the largest possible two-day success ratio that Rahul could have achieved?
1)81/125 2)349/500 3)can not be determined 4)none of these
Hi Puys,
Kindly explain the below.
Sanjog and Rahul took part in a two-day maths contest. At the end both had attempted questions worth 500 marks. Sanjog scored 160 out of 300 attempted on the first day and 140 out of 200 attempted on the second day, so his two-day success ratio was 3/5. Rahul's attempted figures were different from Sanjog's (but with the same two-day total). Rahul had a positive integer marks on each day. For each day Rahul's success ratio was less than Sanjog's. What is the largest possible two-day success ratio that Rahul could have achieved?
1)81/125 2)349/500 3)can not be determined 4)none of these
Hi Puys,
Kindly explain the below.
Sanjog and Rahul took part in a two-day maths contest. At the end both had attempted questions worth 500 marks. Sanjog scored 160 out of 300 attempted on the first day and 140 out of 200 attempted on the second day, so his two-day success ratio was 3/5. Rahul's attempted figures were different from Sanjog's (but with the same two-day total). Rahul had a positive integer marks on each day. For each day Rahul's success ratio was less than Sanjog's. What is the largest possible two-day success ratio that Rahul could have achieved?
1)81/125 2)349/500 3)can not be determined 4)none of these
Rahul got ratios of less than 8/15 on day 1 and 7/10 on day 2. As, the total comes out to be 500, the best thing would be to solve all questions on day 2. So, Rahul will have to solve less than 500 * (7/10) = less than 350 questions i.e. 349 questions. But then, the number of questions he got correct on both days have to be positive integers. So, he has to solve at least 2 questions on day 1 (out of which, he gets 1 right) and the remaining 498 can be solved on day 2 to ensure maximum success.
So, number of marks got on day 2 has to be less than 498 * (7/10) = less than 348.6 questions i.e. 348. So, total comes out to be 349/500.
Option (2)
Is that what you were looking for?
Rahul got ratios of less than 8/15 on day 1 and 7/10 on day 2. As, the total comes out to be 500, the best thing would be to solve all questions on day 2. So, Rahul will have to solve less than 500 * (7/10) = less than 350 questions i.e. 349 questions. But then, the number of questions he got correct on both days have to be positive integers. So, he has to solve at least 2 questions on day 1 (out of which, he gets 1 right) and the remaining 498 can be solved on day 2 to ensure maximum success.
So, number of marks got on day 2 has to be less than 498 * (7/10) = less than 348.6 questions i.e. 348. So, total comes out to be 349/500.
Option (2)
Is that what you were looking for?
Yah the answer is option 2.But I couldnot understand,How did u arrive to the poin that "the best thing would be to solve all questions on day 2. So, Rahul will have to solve less than 500 * (7/10) = less than 350 questions i.e. 349 questions."?Please explain
arabinda sahu SaysYah the answer is option 2.But I couldnot understand,How did u arrive to the poin that "the best thing would be to solve all questions on day 2. So, Rahul will have to solve less than 500 * (7/10) = less than 350 questions i.e. 349 questions."?Please explain
That is because, Rahul has to try to get the maximum number of answers correct. The accuracy has to be marginally less than that of Sanjog on both the days. So, it would be wise to solve maximum possible questions on day 2 as Sanjog has a higher accuracy on that day.
For example suppose Rahul answers all the questions on day one. Now, he has to be slightly less accurate than 8/15. So, he will get a maximum of 500 * (8/15) i.e. 266 answers right. That would be the minimum number you would be getting.
PS: The example does not take into account the 'positive integer only' constraint. It's just for your understanding 😃