先阅读下列材料,然后解答问题:
材料1 从3张不同的卡片中选取2张排成一列,有6种不同的排法,抽象成数学问题就是从3个不同元素中选取2个元素的排列,排列数记为A
32=3×2=6.
一般地,从n个不同元素中选取m个元素的排列数记作A
nm ,
A
nm=n(n-1)(n-2)…(n-m+1)(m≤n).
例:从5个不同元素中选3个元素排成一列的排列数为:A
53=5×4×3=60.
材料2 从3张不同的卡片中选取2张,有3种不同的选法,抽象成数学问题就是从3个元素中选取2个元素的组合,组合数记为C
32=
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmfrac%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3C%2Fmath%3E)
=3.
一般地,从n个不同元素中选取m个元素的组合数记作C
nm ,
C
nm=
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmfrac%3E%3Cmrow%3E%3Cmi%3En%3C%2Fmi%3E%3Cmfenced%3E%3Cmrow%3E%3Cmi%3En%3C%2Fmi%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfenced%3E%3Cmo%3E%E2%8B%AF%3C%2Fmo%3E%3Cmi%3En%3C%2Fmi%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmfenced%3E%3Cmrow%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfenced%3E%3Cmo%3E%E2%8B%AF%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3C%2Fmath%3E)
(m≤n).
例:从6个不同元素中选3个元素的组合数为:
C
63=
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmfrac%3E%3Cmrow%3E%3Cmn%3E6%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E5%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E4%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmo%3E%C3%97%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3C%2Fmath%3E)
=20.
问:(1)从7个人中选取4人排成一排,有多少种不同的排法?
(2)从某个学习小组8人中选取3人参加活动,有多少种不同的选法?