Q. A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair is
answer = 80 ways solution - select 4 pairs from 5 = 5C4 = 5 , then select 1 from a pair = 2C12C12C12C1 = 16 , 165 = 80
please can anyone solve this by using all the 10 shoes , when I solved this , I got 1920. my method-> 10C1= 10 selected 1 shoe, the removed that shoe's pair => no of shoes left = 8 . then selected 1 shoe from remaining 8 => 8C1 and applying this method I got 1086*4 = 1920
The punchline is whether or not you care about the order in which the shoes are removed... or if all you care about are the collection of shoes removed regardless of order. If the shoes are labeled $a,A,b,B,c,C,d,D,e,E$... do you consider having chosen $a,B,D,e$ to be the same or different result as having chosen $B,a,e,D$?
Notice that $1920 = 80\times 4!$
Both are correct answers to different interpretations of the problem. The answer of $80$ is if order did not matter. The answer of $1920$ is if order did matter. To take your answer and change it to the interpretation that order doesn't matter, you can "divide by symmetry"... giving $\dfrac{1920}{4!} = 80$.
When I read the question, I would personally have interpreted it that order should not matter.
As an aside, when it comes to probability it is often a moot point whether you treat order as relevant or not. So long as you are consistent for both your numerator and denominator you will arrive at the same final answer. Here, "What is the probability that if you take four shoes at random that none belong to the same pair?" you can see this as $\dfrac{\binom{5}{4}\times 2^4}{\binom{10}{4}}=\dfrac{8}{21}$ having treated order as irrelevant or as $\dfrac{10\times 8\times 6\times 4}{10\times 9\times 8\times 7}=\dfrac{8}{21}$ having treated order as relevant. Both answers are of course equal, just differently represented.