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

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.

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!

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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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