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# A bibliophile plans to put a total of seven books on her marble shelf

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Manager
Joined: 14 Oct 2015
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A bibliophile plans to put a total of seven books on her marble shelf  [#permalink]

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19 Jul 2017, 05:38
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2
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Difficulty:

35% (medium)

Question Stats:

73% (02:44) correct 28% (02:44) wrong based on 57 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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Re: A bibliophile plans to put a total of seven books on her marble shelf  [#permalink]

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19 Jul 2017, 05:48
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jedit wrote:
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

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.

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Re: A bibliophile plans to put a total of seven books on her marble shelf  [#permalink]

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19 Jul 2017, 05:53
jedit wrote:
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

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
D
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A bibliophile plans to put a total of seven books on her marble shelf  [#permalink]

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19 Jul 2017, 05:56
jedit wrote:
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

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  [#permalink]

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20 Jul 2017, 08:51
jedit wrote:
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

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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Re: A bibliophile plans to put a total of seven books on her marble shelf  [#permalink]

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20 Jul 2017, 09:21
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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Re: A bibliophile plans to put a total of seven books on her marble shelf   [#permalink] 20 Jul 2017, 09:21
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