Today, faced with this problem: Consider $ 1, 2, \ dots, n $ of all permutations, a total of $ n $ a!. Now specify the order in which the relative number of $ k $, ask qualifying arrangement How many?
analysis
Without loss of generality, designated $ 1, 2, \ dots, k $ $ k which is the number of the relative order of $ $ 1, 2, \ dots, k $. For example words, set $ n = 5, k = 2 $, then $ (1, 2, 3, 4, 5) $, $ (1, 4, 3, 2, 5) $, $ (1, 5, 3, 4, 2) are qualified $ and $ (3, 2, 5, 1, 4) $, $ (2, 1, 4, 3, 5) $ have failed.
Consider the following arrangement configuration procedure:
first $ 1, 2, \ dots, k $ sequentially lined up, and then $ k + 1, \ dots, n $ inserted one by one into the sequence. Still $ n = 5, k = 2 $ , for example, can be $ (1, 4, 3, 2, 5) $ imagine this arrangement is obtained by the following four steps:
$ (1, 2) \ to (1, 3, 2) \ to (1, 4, 3, 2) \ to (1, 4, 3, 2, 5) $.
Not difficult to prove: the arrangement are so constructed to meet the conditions, and all meet the conditions of the arrangement can be so constructed.
When there is inserted $ k + 1 $ $ k + 1 $ positions available, is inserted with a $ k + 2 $ $ k + 2 $ positions available when ......, has n-$ $ $ n-th position is inserted $ optional. Thus the arrangement can be seen to meet the conditions of total $ (k + 1) (k + 2) \ dots (n) = n! / K! $ A.
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$ 1 to $ $ all permutations of n-$ meet $ x $ ($ k <x \ le n $) ranked $ 1, 2, \ dots, arranged before the number of k $?
Followed ideas arranged above configuration, difficult to draw a total of $ k! N! / (K + 1)! = N! / (K + 1) $.
Another idea may also be used to describe the amount required: $ X $ enumerated position in the arrangement, there are
$ n / (k + 1) = \ sum_ {i = 1} ^ {n - k} \! binom {n - i} {k } k! (n - k - 1)! $, i.e.
$ \ sum_ {i = 1} ^ {n - k} \ binom {n - i} {k} = n /! ((k + 1)! ( nk-1)!) = \ binom {n} {k + 1} $.
Indicators do substitution, so $ j = n - i $, $ m = n - 1 $, to give
$ \ sum_ {j = k} ^ {m} \ binom {j} {k} = \ binom {m + 1 } {k + 1} $.