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Saikat4292
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D01-39 26 May 2025, 23:02
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D01-39 25 Jul 2025, 07:31
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D01-39 14 Aug 2025, 04:45
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D01-39 15 Aug 2025, 13:12
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D01-39 16 Nov 2025, 00:15
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D01-39 16 Nov 2025, 02:00
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D01-39 30 Nov 2025, 05:12
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I did not quite understand the solution. Why should we consider him bringing 5 small books? Why not 4 small books + 1 large books only and 3 small books + 2 large books?
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D01-39 30 Nov 2025, 05:18
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sunshineeee
I did not quite understand the solution. Why should we consider him bringing 5 small books? Why not 4 small books + 1 large books only and 3 small books + 2 large books?
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The rule says at most 1 large. That means 0 large allowed or 1 large allowed.

So the only valid selections are:

• 5 small and 0 large
• 4 small and 1 large

The option “3 small + 2 large” is not allowed because that uses 2 large books, which violates “at most 1”.
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D01-39 25 Dec 2025, 11:58
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D01-39 01 Jan 2026, 07:02
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D01-39 01 Feb 2026, 06:18
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I did not quite understand the solution. How did we find the number 56 and 140?
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D01-39 01 Feb 2026, 06:32
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Globalstudent114
I did not quite understand the solution. How did we find the number 56 and 140?
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C stands for combinations.

 \(C^5_8=\frac{8!}{5!(8-5)!}= \frac{8!}{5!3!}=56\)

\(C^4_8*C^1_2=\frac{8!}{4!(8-4)!} * \frac{2!}{1!(2-1)!}= \frac{8!}{4!4!} * \frac{2!}{1!1!}=140\)


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D01-39 17 Mar 2026, 10:32
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