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18 Feb 2021, 06:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (02:09) correct
38%
(02:05)
wrong
based on 161
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There are 12 books on Algebra and Calculus in our library, the books of the same subject being different. If the number of selection each of which consists of 3 books on each topic is greatest then the number of books of Algebra and Calculus in the library are respectively:
(A) 3 and 9
(B) 4 and 8
(C) 5 and 7
(D) 6 and 6
(E) 6 and 7
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 3 and 9
(B) 4 and 8
(C) 5 and 7
(D) 6 and 6
(E) 6 and 7
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
18 Feb 2021, 13:26
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Bunuel
Split 12 into 6 and 6 and that would be the answer, this gives us \(6C3 * 6C3 = (\frac{6*5*4}{3!})^2 = 20^2 = 400\) selections. The next closest option is C which is \(5C3 * 7C3 = \frac{5*4*3}{3!} * \frac{7*6*5}{3!} = 10*35 = 350\). E doens't add up to 12 so it is rejected too.
Generally, two medium numbers in multiplication are larger than one large and one small number, and a similar optimization concept applies here.
Ans: D
14 Apr 2025, 22:16
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Step 1: Understand the problem
Step 2: Identify the relationship
Since \( a+c=12\), for any selection, c is determined by a. So we need to maximize \(\binom{a}{3} \times \binom{12-a}{3}\)
Step 3: Use symmetry
The combination function \(\binom{n}{3}\) increases as n increases and decreases as n decreases, but the product \(\binom{a}{3} \times \binom{12-a}{3}\) is typically maximized when a and c are as equal as possible. This is because the values of \(\binom{a}{3}\) and \(\binom{c}{3}\) will be maximized when a and c are close to each other.
Step 4: Check the cases for balanced values of a and c
Given \(a+c=12\), the two closest possible values for a and c are:
Thus, the correct answer is:
(D) 6 and 6
- There are 12 books in total: some on Algebra and some on Calculus.
- We are selecting 3 books from Algebra and 3 books from Calculus.
- The goal is to maximize the number of selections.
- The number of ways to select 3 books from Algebra is \(\binom{a}{3}\), where a is the number of Algebra books.
- The number of ways to select 3 books from Calculus is \(\binom{c}{3}\), where c is the number of Calculus books.
- We need to maximize \(\binom{a}{3} \times \binom{c}{3}\) under the constraint that \(a+c=12\).
Step 2: Identify the relationship
Since \( a+c=12\), for any selection, c is determined by a. So we need to maximize \(\binom{a}{3} \times \binom{12-a}{3}\)
Step 3: Use symmetry
The combination function \(\binom{n}{3}\) increases as n increases and decreases as n decreases, but the product \(\binom{a}{3} \times \binom{12-a}{3}\) is typically maximized when a and c are as equal as possible. This is because the values of \(\binom{a}{3}\) and \(\binom{c}{3}\) will be maximized when a and c are close to each other.
Step 4: Check the cases for balanced values of a and c
Given \(a+c=12\), the two closest possible values for a and c are:
- \(a=6, c=6\)
- a=5, c=7 or a=7, c=5
Thus, the correct answer is:
(D) 6 and 6
