从5双不同鞋号中选4只一共10!/(4!6!)种选法
2只配成1双5!/(1!4!)*(8!/2!6!)-5!/(2!3!) (-5!/(2!3!)是因为重复了1次)
至少配成1双的概率[5!/(4!1!)*8!/(6!2!)*4-5!/(2!3!)]/10!(4!6!)=[(5*28)-10]/[(10*9*8*7)/(4*3*2)]=130/210=13/21
从5双不同鞋号中选4只一共10!/(4!6!)种选法
2只配成1双5!/(1!4!)*(8!/2!6!)-5!/(2!3!) (-5!/(2!3!)是因为重复了1次)
至少配成1双的概率[5!/(4!1!)*8!/(6!2!)*4-5!/(2!3!)]/10!(4!6!)=[(5*28)-10]/[(10*9*8*7)/(4*3*2)]=130/210=13/21