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Carrotparrot
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The number of ways in which a captain and a vice-captain can be Updated on: 24 Oct 2024, 09:14
Originally posted by Carrotparrot on 24 Oct 2024, 08:47.
Last edited by Bunuel on 24 Oct 2024, 09:14, edited 3 times in total.
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72% (02:54) correct 28% (03:25) wrong based on 99 sessions

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The number of ways in which a captain and a vice-captain can be nominated from all the students in Class X is 590 more than the number of ways in which a captain and a vice-captain can be nominated from all the students in Class Y. No student can be nominated as both a captain and a vice-captain. If there are 10 students more in Class X than in Class Y, how many students are there in Class X ?

A) 15
B) 20
C) 25
D) 35
E) 40
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The number of ways in which a captain and a vice-captain can be 24 Oct 2024, 08:55
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Moving to a better sub-forum for PS questions.
What is the source of this question Carrotparrot?

Also, can you check to make sure there are no missing Class items? it seems there is something missing in the question but that may be me.
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The number of ways in which a captain and a vice-captain can be 24 Oct 2024, 09:23
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Assuming number of students in class X is n.

nP2 = 590 + (n-10)P2
n!/(n-2)! = 590+(n-10)!/(n-12)!
n(n-1) = 590+(n-10)(n-11)
n^2 -n = 590+n^2 -21n+110
-n = 590-21n+110
20n = 700
n = 700/20
n = 35
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The number of ways in which a captain and a vice-captain can be 29 Oct 2024, 01:46
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Carrotparrot
The number of ways in which a captain and a vice-captain can be nominated from all the students in Class X is 590 more than the number of ways in which a captain and a vice-captain can be nominated from all the students in Class Y. No student can be nominated as both a captain and a vice-captain. If there are 10 students more in Class X than in Class Y, how many students are there in Class X ?

A) 15
B) 20
C) 25
D) 35
E) 40
Hi,
Let # of students in class Y= a
Thus, # of students in class X = a+10

Given that,
# of ways we can pick 2 different students for Cap. & Vice Cap. from X = 590 + # of ways we can pick 2 different students for Cap. & Vice Cap. from Y
Therefore,
# of ways we can pick 2 different students for Cap. & Vice Cap. from X = Picking 2 students from (a+10) students = \( C^2_(a+10) \) = \([ (a+10)(a+9)]/2 \)

# of ways we can pick 2 different students for Cap. & Vice Cap. from Y = Picking 2 students from (a) students = \( C^2_a \) = \([(a)(a-1)]/2\)

\([(a+10)(a+9)]/2\) = 590 + \([(a)(a-1)]/2\)
On simplifying, we get,
20a = 500
a= 25
Thus, # of students in class X = a+10 = 25+10 = 35 Ans
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The number of ways in which a captain and a vice-captain can be 19 Jan 2025, 05:11
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There's a problem with this simplification. The equation we should get in the end is: 10a = 545 which gives us a=54,5
This can't be since a represents the number of students in class Y. So I'm guessing there's something wrong with the question.
zlishz
Carrotparrot
The number of ways in which a captain and a vice-captain can be nominated from all the students in Class X is 590 more than the number of ways in which a captain and a vice-captain can be nominated from all the students in Class Y. No student can be nominated as both a captain and a vice-captain. If there are 10 students more in Class X than in Class Y, how many students are there in Class X ?

A) 15
B) 20
C) 25
D) 35
E) 40
Hi,
Let # of students in class Y= a
Thus, # of students in class X = a+10

Given that,
# of ways we can pick 2 different students for Cap. & Vice Cap. from X = 590 + # of ways we can pick 2 different students for Cap. & Vice Cap. from Y
Therefore,
# of ways we can pick 2 different students for Cap. & Vice Cap. from X = Picking 2 students from (a+10) students = \( C^2_(a+10) \) = \([ (a+10)(a+9)]/2 \)

# of ways we can pick 2 different students for Cap. & Vice Cap. from Y = Picking 2 students from (a) students = \( C^2_a \) = \([(a)(a-1)]/2\)

\([(a+10)(a+9)]/2\) = 590 + \([(a)(a-1)]/2\)
On simplifying, we get,
20a = 500
a= 25
Thus, # of students in class X = a+10 = 25+10 = 35 Ans
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The number of ways in which a captain and a vice-captain can be 19 Jan 2025, 05:54
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The explanation you're referring to misses out on multiplying aC2 by 2, after selecting 2 students there are 2 ways students can get a particular post.

Below is how I solved,

\(x(x-1) = y(y-1) + 590\)

Since \(x = y+10\)

\(y+10(y+9) = y(y-1) + 590\)

\(y^{2}+19y+90 = y^{2} - y + 590\)

\(20y = 500\)

y = 25
x = 35
Naoufal
There's a problem with this simplification. The equation we should get in the end is: 10a = 545 which gives us a=54,5
This can't be since a represents the number of students in class Y. So I'm guessing there's something wrong with the question.
zlishz
Carrotparrot
The number of ways in which a captain and a vice-captain can be nominated from all the students in Class X is 590 more than the number of ways in which a captain and a vice-captain can be nominated from all the students in Class Y. No student can be nominated as both a captain and a vice-captain. If there are 10 students more in Class X than in Class Y, how many students are there in Class X ?

A) 15
B) 20
C) 25
D) 35
E) 40
Hi,
Let # of students in class Y= a
Thus, # of students in class X = a+10

Given that,
# of ways we can pick 2 different students for Cap. & Vice Cap. from X = 590 + # of ways we can pick 2 different students for Cap. & Vice Cap. from Y
Therefore,
# of ways we can pick 2 different students for Cap. & Vice Cap. from X = Picking 2 students from (a+10) students = \( C^2_(a+10) \) = \([ (a+10)(a+9)]/2 \)

# of ways we can pick 2 different students for Cap. & Vice Cap. from Y = Picking 2 students from (a) students = \( C^2_a \) = \([(a)(a-1)]/2\)

\([(a+10)(a+9)]/2\) = 590 + \([(a)(a-1)]/2\)
On simplifying, we get,
20a = 500
a= 25
Thus, # of students in class X = a+10 = 25+10 = 35 Ans
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The number of ways in which a captain and a vice-captain can be 26 Jan 2025, 10:42
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I solved that one quite easily plugging the answers.

The number of ways having a captain and a vice captain for a class with n students is n(n-1).
We also know that a class have 590 more ways than another to do so, so we can deduce that n(n-1)>590. That eliminates answers A and B. We can also eliminate answer C by approximating n(n-1) to n^2, which gives us 25^2 = 625, way too close to 590 to be the correct answer.

We then plug answer D, computing 35*34 and 625 - 25 (as 25*25 - 25 = 25*24) and find that it is the right one.
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The number of ways in which a captain and a vice-captain can be 26 Jan 2025, 11:14
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You messed up your numbers pretty hard brother. The answer with your numbers is Y = 54,5 and X = 64,5.

Unless there is half human beings, the solution doesn't work.

Equation is : 590 + (Y*(Y-1))/2 = ((Y+10)*(Y+9))/2
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The number of ways in which a captain and a vice-captain can be 27 Jan 2025, 02:21
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Maybe my english is not great and I did not express myself crealy, but when I was solving the problem, I found the correct answer so let me try to re explain.
I plugged the answer to find the correct one. I just eliminated the three first ones by logical eliminiation.
Then, if you take X = 35, you have number of ways captain and vice captains are elected is 35*34 = 1190.
If X = 35, Y = 25, so number of ways captain and vice captains are elected is 25*24 = 600. The difference between the two is 590, so the answer choice D (X=35) satisfies the contraints of the problem.


Let me know if you had already understood that and so if my answer is still not correct in the reasoning.
Montyrafy
You messed up your numbers pretty hard brother. The answer with your numbers is Y = 54,5 and X = 64,5.

Unless there is half human beings, the solution doesn't work.

Equation is : 590 + (Y*(Y-1))/2 = ((Y+10)*(Y+9))/2
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