# Find the number of ways an odd number of objects can be chosen

## Homework Statement

Find the number of ways in which an odd number of objects can be chosen from n objects.

## The Attempt at a Solution

We have 2 cases-

1) n is even

nC1 + nC3 + nC5 + ...... nCn-1

2) n is odd

nC1 + nC3 + nC5 + ...... nCn

How do I find out the sum?

tiny-tim
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Hi Abdul! Hint: for n odd, it's easy! (if you can't see it immediately, try a few examples … in fact, trying a few examples is often a good idea anyway)

I used binomial expansions and surprisingly, got both of them as 2n-1.

tiny-tim
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Since that's half of 2n, that's not at all surprising for odd n …

can you see why? Its obvious for odd n, what about even n?

tiny-tim
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sorry, are you saying you can see it's obvious for odd n, or not? Yeah I can see it's obvious for odd n.
In the expansion of (1+x)n, the number of terms are n+1 = even number of terms , n being odd.

Of these even number of terms, half correspond to choosing odd number of objects and half correspond to choosing an even number of objects. Sum of any half = 2n-1

If n is even ( as in the case 1 ), how do you justify 2n-1 (the answer I got)?

tiny-tim
Homework Helper
Hi Abdul! Of these even number of terms, half correspond to choosing odd number of objects and half correspond to choosing an even number of objects. Sum of any half = 2n-1

Yes, that's right. But I can't see an obvious way of proving it for even n

(maybe someone else can? )

though I can see a neater maths way … consider (1 + 1)n + (1 - 1)n verty
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