Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

question on binomial theory..?

hi every1, can any1 prove that (n+1 C r+1) = n C r + (n C r+1)

or explain to me where did this relation come from?

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  • 1 decade ago
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    First, let's state the meaning of nCr.

    Consider n objects. How many ways are there to take r objects from these n objects?

    Well, it is called nCr.

    Now, with resort to this definition and a little mental idea we can prove the given statement.

    It reads that the number of ways to choose (r+1) objects from among (n+1) ones is equal to the number of ways to choose r objects from n objects, plus the number of ways to choose (r+1) objects from among n ones.

    Idea: divide the n+1 objects into two groups; one containing one element and the other the rest of the n elements.

    Now, the left side (n+1)C(r+1) can be decomposed into the number of ways to choose (r+1) elements from the n elements in the second group above, and the number of ways to choose (r+1) elements from both groups, which means taking the first group unique element and r elements from the second group of n elements. This decomposition translated to mathematical terms reads nC(r+1) + nCr.

  • 1 decade ago

    (n +1)C (r+1) means the number of combinations of (n +1) things taken (r+1) at a time. Each combination contains (r+1) things.

    Let us reserve or remove one thing from the (n +1) things so that we are left with n things.

    The number of combinations of (n) things taken (r+1) at a time is nC (r+1).Each combination contains (r+1) things.

    The above number combinations are done leaving a particular thing.

    If that thing were added we could have more number of combinations.

    To find that number of combinations that we have lost, among the n things we take r at a time, so that the number of combinations is nCr. Each combination contains only r things (with out the thing we have reserved).

    We add the left thing so that each combination will have (r+1) things.

    (n +1)C (r+1) = nC (r+1) +nCr.

  • yes i can. we will start with n=10, r=5

    (10+1) choose (5+1) = 462

    10 choose 5= 252

    10 choose 6 = 210

    210+252= 462

    I am assuming you know how to figure out factorials. I hope this helps.

  • 1 decade ago

    (n C r) + (n C r+1) = n!/r!(n-r)! + n!/(r+1)!(n-r-1)!

    = n!(r+1)/r!(r+1)(n-r)! + n!(n-r)/(r+1)!(n-r)(n-r-1)!

    (times (r+1)on the above and below part of 1st term,

    times (n-r)on the above and below part of 2nd term)

    = n!(r+1)/(r+1)!(n-r)! + n!(n-r)/(r+1)!(n-r)!

    = n![(r+1)+(n-r)]/(r+1)!(n-r)!

    = n!(n+1)/(r+1)!(n-r)!

    = (n+1)!/(r+1)![(n+1)-(r+1))]!

    = (n+1 C r+1)

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  • evariste galois

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