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number of ways to pick k distinct objects from n objects Watch

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    for example, the number of ways to pick 2 distinct numbers from the set {1, 2, 3, 4, 5} (so making k = 2, & n = 5) is:

    1, 2
    1, 3
    1, 4
    1, 5
    2, 3
    2, 4
    2, 5
    3, 4
    3, 5
    4, 5

    That's 10 ways all together.

    How do I generalise this to picking k distinct objects in n?
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    This is simply the binomial coefficient nCk , which can also be written as n!/ [ k! * (n-k)! ]

    Hope it helps. Peace.
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    (Original post by WhiteGroupMaths)
    This is simply the binomial coefficient nCk , which can also be written as n!/ [ k! * (n-k)! ]

    Hope it helps. Peace.
    yes it did, thanks!
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    (Original post by Dr Ben)
    for example, the number of ways to pick 2 distinct numbers from the set {1, 2, 3, 4, 5} (so making k = 2, & n = 5) is:

    1, 2
    1, 3
    1, 4
    1, 5
    2, 3
    2, 4
    2, 5
    3, 4
    3, 5
    4, 5

    That's 10 ways all together.

    How do I generalise this to picking k distinct objects in n?
    The first object selected can be any of the n given, the second object any of the remaining (n-1), the third object, any of the remaining (n-2) and so on until the kth object is any of the remaining (n-k+1) giving a total of nx(n-1)x(n-2)x(n-3)x...x(n-k+1)
    which may be written as  \frac {n!}{(n-k)!}
    But since we are not interested in the order in which the objects are chosen we must take into account that each different selection will be repeated k! times so final; answer is  \frac {n!}{(n-k)!\times k!}
 
 
 
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