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jedit
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A bibliophile plans to put a total of seven books on her marble shelf 19 Jul 2017, 05:38
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55% (hard)

Question Stats:

67% (02:54) correct 33% (03:06) wrong based on 421 sessions

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A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

A - 441
B - 1225
C - 1666
D - 1715
E - 1820
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D
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A bibliophile plans to put a total of seven books on her marble shelf 19 Jul 2017, 05:48
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A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

A - 441
B - 1225
C - 1666
D - 1715
E - 1820
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We are choosing from 7 Antiquity books and 7 Post-modernism books 7 books so that there are at least 4 from Antiquity, and 1 on Post-Modernism.

1. 4 from Antiquity, and 3 on Post-Modernism: \(C^4_7*C^3_7=35*35=35^2\)

2. 5 from Antiquity, and 2 on Post-Modernism: \(C^5_7*C^2_7=21*21=21^2\)

3. 6 from Antiquity, and 1 on Post-Modernism: \(C^6_7*C^1_7=7*7=7^2\)

Total = 35^2 + 21^2 + 7^2 = 1715.

Answer: D.
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A bibliophile plans to put a total of seven books on her marble shelf 19 Jul 2017, 05:53
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A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

A - 441
B - 1225
C - 1666
D - 1715
E - 1820
Show more



Hi

We are looking at ATLEAST 4 of A and 1 of P....
Following ways..
1) 7C4*7C3 =\(\frac{7!}{4!3!}^2=35*35=1225\)
2) 7C5*7C2.=\(\frac{7!}{5!2!}^2=21*21=441\)
3)7C6*7C1..7*7=49

Total ways= 1225+441+49=1715
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A bibliophile plans to put a total of seven books on her marble shelf 19 Jul 2017, 05:56
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A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

A - 441
B - 1225
C - 1666
D - 1715
E - 1820
Show more

A = Antiquity, P = Post-modernism.

The possible cases are:
4A + 3P: 7c4 * 7c3 = 35 * 35 = 1225
5A + 2P: 7c5 * 7c2 = 21 * 21 = 441
6A + 1P: 7c6 * 7c1 = 7 * 7 = 49

Total no. of ways = 1715. Ans - D.
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A bibliophile plans to put a total of seven books on her marble shelf 20 Jul 2017, 08:51
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A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

A - 441
B - 1225
C - 1666
D - 1715
E - 1820
Show more
I solved the same way as others, and just want to highlight a time saver: At each stage, this question involves the identity

\(_nC_k = {_n}C_{(n-k)}\)

In other words,

\(_7 C _4 = {_7}C _3\)
\(_7 C _5 = {_7}C _2\)
\(_7 C _6 = {_7}C _1\)

Basically, selecting which \(k\) objects DO get chosen is the same as selecting which \((n - k)\) objects DO NOT get chosen.

Once you compute the first combination in each pair, you're done; it's exactly the same value as the second combination in the pair.

The point is obvious for people who are really fluent in combinations. I'm not. I don't like them. :-) So knowing that identity saved me a decent chunk of time here. Cheers!
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A bibliophile plans to put a total of seven books on her marble shelf 20 Jul 2017, 09:21
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Can somebody explain me where I am getting wrong. I did like this
7C4*7C1*9C2 (4 chosen from A, 1 chosen from P and 2 from remaining all)

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A bibliophile plans to put a total of seven books on her marble shelf 27 Dec 2022, 03:19
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I also did it the same way as Kishorpalleti and also don't understand why I am wrong

We must have at least 4 of Antiquity hence we can use 7C4 to choose the 4 that we will use
We must have at least 1 from Post Modernism so we can use 7C1 to choose the one book there

This leaves 2 book choices to make and 9 books in total left (3 from Antiquity and 6 on PM) so we can use 9C2 because any combination of the books works
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A bibliophile plans to put a total of seven books on her marble shelf 16 Sep 2025, 03:49
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sydqur kishorpalleti
It's because this combination leaves significant room for overcounting.

The core problem is that this method treats the selection of the initial 5 books (4 Antiquity, 1 Post-modernism) and the final 2 books as two separate, independent steps. However, the identity of the final group of 7 books is what matters, not the order or steps in which they were chosen.

The correct way to solve problems with "at least" conditions is to partition the total possibilities into a series of mutually exclusive (non-overlapping) cases. Each case represents a unique final composition of the 7 books that satisfies the constraints.

The possible compositions are defined by the number of Antiquity books (A) and Post-modernism books (P) selected, where A+P=7, A≥4, and P≥1.

Case 1: 4 Antiquity and 3 Post-modernism.

Number of ways: C(7,4)×C(7,3)=35×35=1225

Case 2: 5 Antiquity and 2 Post-modernism.

Number of ways: C(7,5)×C(7,2)=21×21=441

Case 3: 6 Antiquity and 1 Post-modernism.

Number of ways: C(7,6)×C(7,1)=7×7=49

By using this method, each unique final group of 7 books (e.g., a specific set of 5 Antiquity and 2 Post-modernism books) is calculated only once within its specific case. Summing the results of these distinct cases gives you the correct total number of ways:

1225+441+49=1715
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