Q. Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible? A. 10 B. 21 C. 210 D. 420 E. 1,260
Could someone please explain the solution to this?
Thanks, Rohit.
i think the answer is E- 1260:hurray:
The explanation goes like this
choosing 5 contestants out of 7 first, then one out of 5, then one out of 4 and lastly one out of 3.
So it comes out to be C(7,5) x C(5,1) x C(4,1) x C(3,1) = 1260
my pick will be 7P3 ie option C. Pick three people out of 7 and arrange in 3! ways. alternatively, the first prize can go in 7 different ways, the second in 6 possible ways and third in 5 possible ways, creating a unique arrangement and total number would be a product of three.
hmmm, when ur picking, i think u shud u Combination rather than Permutation.
So it shud be 7C3 and even that is incorrect coz out of 7 only 5 are the final contestants.
the theoritical limit to the answer is 7p3, whatever the hell you do in the world you cannot have more than 7p3 ways to pick and arrange 3 people. I just looked it this way that in the end there would be three guys squatting out there on the podium with colored ribbons around their necks and i have to do that with 7 to start with. So, in between whatever I do has no effect whatsoever on the number of arrangements ( like pick 6 out of 7 first and then 4 out of those six and then pick 3 out of those 4).........in its simplest form say you have 7 people and only 1 prize, i'll be dashed if there are more than 7 ways to do so. You are confusing it with the number of processes to arrive at that 1 person, the question asks the outcomes possible and not the ways in which you can get the outcome 😃
the theoritical limit to the answer is 7p3, whatever the hell you do in the world you cannot have more than 7p3 ways to pick and arrange 3 people. I just looked it this way that in the end there would be three guys squatting out there on the podium with colored ribbons around their necks and i have to do that with 7 to start with. So, in between whatever I do has no effect whatsoever on the number of arrangements ( like pick 6 out of 7 first and then 4 out of those six and then pick 3 out of those 4).........in its simplest form say you have 7 people and only 1 prize, i'll be dashed if there are more than 7 ways to do so. You are confusing it with the number of processes to arrive at that 1 person, the question asks the outcomes possible and not the ways in which you can get the outcome :)
sounds interesting n ur funda seems logical as well!!
(1)Two identical taps fill 2/5 of a tank in 20 minutes. When one of the taps goes dry in how many minutes will the remaining one tap fill the rest of the tank ?
1) 5 minutes 2) 10 minutes 3) 15 minutes 4) 20 minutes 5) None of the above
(2)What is the greatest value of a positive integer n such that 3n is a factor of 1815? 1) 15 2) 18 3) 30 4) 33 5) 45
Hi, I did not understood how you solved 2nd question [Take the Prime Factorization of 1815; 1815 = 3 x 5 x 11 x 11 If you can the options, 33 x 3 = 121. This is the greatest value of n in the list for which 3n will be a factor of 1815.}
ONE MORE (3) If r = (3p + q)/2 and s = p - q, for which of the following values of p would r2 = s2(r square=s square)?
1) 1q/5 2) 10 - 3q/2 3) q - 1 4) 3q 5) 9q/2 - 9
answer is P = q/5 (3p+q)^2 = (2p-2q)^2 on expangin and solving u get 5p^2 + 14pq - 3q^2 = 0 5p(p+3q)-q(p+3q) = 0 (5p-q)(p+3q)=0 solving we get p = q/5 it was jus basic solving of two sides of equation.
Hi Aziz Pardon my lack of multi-tasking skills. I got the answer but ended up giving the wrong logic in my previous post. I have corrected the same now. Thank you for bringing this to my notice
Cheers...Raj
i still couldnt get the answer... could u tell me how u did it my answer doesnt match ane of the given ones
Hi Vamsi The number you have is 1815. Take the Prime factorization 1815 = 3 x 5 x 11 x 11 The other factors can be 15, 33, 55, 121, 165, 363 etc
When you check your answer options only look for options from this list. Of all the options that you have, 15 and 33 are the only 2 options which can be derived from the prime factorization of 1815. However 15 and 33 will be values of x. You need to check the divisibility with 3x i.e. 45 and 99.
Hence the correct answer is 33.
hey coach.. even i narrowed doen to 15 and 33 jus by lokoing at the options... i think he was asking for value of n and if n is 33... 3n is 99.. and as far as i learnt maths i don think 99 is a factor of 1815... the answer according to me is not in the given options
Q. Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible? A. 10 B. 21 C. 210 D. 420 E. 1,260
Could someone please explain the solution to this?
Thanks, Rohit.
:) :) :)
the theoritical limit to the answer is 7p3, whatever the hell you do in the world you cannot have more than 7p3 ways to pick and arrange 3 people. I just looked it this way that in the end there would be three guys squatting out there on the podium with colored ribbons around their necks and i have to do that with 7 to start with. So, in between whatever I do has no effect whatsoever on the number of arrangements ( like pick 6 out of 7 first and then 4 out of those six and then pick 3 out of those 4).........in its simplest form say you have 7 people and only 1 prize, i'll be dashed if there are more than 7 ways to do so. You are confusing it with the number of processes to arrive at that 1 person, the question asks the outcomes possible and not the ways in which you can get the outcome :)
WELL!!! i would like to stress upon the fact that the question says 5 out of 7 are the finalist!!!! if we use 7P3, then we are taking ALL 7 into consideration and NOT 5 as the question demands!!!
So i m deadly sure that my answer is correct. Prove me WRONG if possible....
1. What is the probability that a leap year will have 53 Sundays
2. What is the sum of the first 30 terms of the following series The series has the first 2 terms 2, 3. Each subsequent term is the sum of all previous terms.
Cheers...Raj
Hi Raj,
The answers to your questions.
1. 2/7
If a leap year starts with Su, then it will end with Mo, (52*7=)364 + 2; so 52 complete weeks + Su + Mo If it starts with a Sa; there is another possibility for a 53rd Su. If the year starts with remaining days then there will be only 52 Su.
2. 5*( 2^28 )
The easy way to solve these questions is try to calculate for a few terms and observe the pattern.
Sum of first 2 terms in the sequence = 2+3 = 5 = 5*(2^0) Sum of first 3 terms = 2+3+5 = 10 = 5*(2^1) from here on when you add the next term it will be 2 times the previous sum; i.e sum of first 4 terms = 2 * (sum of first 3 terms) and so on...
so, sum of 3 terms = 10 = 5 * (2^1) = 5 * (2^ (3-2)) sum of 4 terms = 10 = 5 * (2^2) = 5 * (2^ (4-2))
How many 4 digit numbers divisible by 5 can be formed by digits 0, 1, 2, 3, 4, 5, 6 and 6.
Is there a direct formula for this? I tried solving this by considering two methods and am getting different results with each method :(.
Hi sarabji_dua, i dont know about any shortcut as such............ is repitition is allowed or not is not mentioned......... considering the numbers to be 0,1,2,3,4,5,6
1) if repitition is allowed first position can be filled in 6 ways(exclude 0), 2nd in 7 ways , 3rd in 7 ways and 4th in two ways (0,5 as only this two digit in the end will make the number divisible by 5) so total ways = 6*7*7*2=588
case 2) if ripitition is not allowed
here there will be two cases one with last digit as 0 and one with last digit as 5 i) if last digit is 0........ then fist position can be filled in 6 ways, 2nd in 5 ways, 3rd in 4 ways and 4th is zero so total ways = 6*5*4*1=120 ii) last digit is 5 then 1st position can be filled in 5 ways(0 excluded), 2nd in 5 ways, and 3rd in 4 ways total ways =5*5*4=100
Hi sarabji_dua, i dont know about any shortcut as such............ is repitition is allowed or not is not mentioned......... considering the numbers to be 0,1,2,3,4,5,6
1) if repitition is allowed first position can be filled in 6 ways(exclude 0), 2nd in 7 ways , 3rd in 7 ways and 4th in two ways (0,5 as only this two digit in the end will make the number divisible by 5) so total ways = 6*7*7*2=588
case 2) if ripitition is not allowed
here there will be two cases one with last digit as 0 and one with last digit as 5 i) if last digit is 0........ then fist position can be filled in 6 ways, 2nd in 5 ways, 3rd in 4 ways and 4th is zero so total ways = 6*5*4*1=120 ii) last digit is 5 then 1st position can be filled in 5 ways(0 excluded), 2nd in 5 ways, and 3rd in 4 ways total ways =5*5*4=100
so here total ways = 120+100=200
hope it helps.......... BTW whats the answer....
Hey Inder,
OA is 588 but i have a feeling thats wrong. Firstly in such questions where nothing is mentioned, we usually don't consider repetition. Another thought, you are neglecting that the question provides two 6 digits (question says 0,1,2,3,4,5,6 and 6). I tried to solve it and am getting a 249 (without repetition).
OA is 588 but i have a feeling thats wrong. Firstly in such questions where nothing is mentioned, we usually don't consider repetition. Another thought, you are neglecting that the question provides two 6 digits (question says 0,1,2,3,4,5,6 and 6). I tried to solve it and am getting a 249 (without repetition).
