解由n!=n（n-1）（n-2）（n-3）.×2×1
即n!=n（n-1）（n-2）（n-3）!
=n（n-1）（n-2）×（n-3）!
故n!/(n-3）!
=n（n-1）（n-2）×（n-3）!/(n-3）!
=n（n-1）（n-2）.
=1,=2乘以1,3=3乘以2乘以1,=4乘以3乘以2乘以1
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