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

  • Thread starter zorro
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  • #1
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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

I think adding them will give the answer.
How do I find out the sum?
 

Answers and Replies

  • #2
tiny-tim
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Hi Abdul! :smile:

Hint: for n odd, it's easy! :wink:

(if you can't see it immediately, try a few examples … in fact, trying a few examples is often a good idea anyway)
 
  • #3
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I used binomial expansions and surprisingly, got both of them as 2n-1.
 
  • #4
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? :wink:
 
  • #5
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Its obvious for odd n, what about even n?
 
  • #6
tiny-tim
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sorry, are you saying you can see it's obvious for odd n, or not? :confused:
 
  • #7
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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)?
 
  • #8
tiny-tim
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Hi Abdul! :wink:
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. :smile:

But I can't see an obvious way of proving it for even n

(maybe someone else can? :smile:)

though I can see a neater maths way … consider (1 + 1)n + (1 - 1)n :wink:
 
  • #9
verty
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Last edited:

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