Variation Of Euler's Letter Problem
Say, there are $n$ letters and $n$ envelops. In how many ways can we assign the letters to the envelops so that $(1).$ exactly $k$ letters are in the wrong envelops and $(2).$ at most $k$ letters are in the wrong envelops?
One one thing is neutral in the universe, that is $0$.
Re: Variation Of Euler's Letter Problem
($1$)$\binom{n}{n-k}d_{k}$.Where $d_{k}$ is the derangement of $k$ letters in $k$ envelops.
($2$)This is the sum of the numbers when no letters are in wrong envelop$+$exactly $2$ letters are in wrong envelops+$3$ letters+.....+$k$ letters are in wrong envelops.
Am I correct .Please inform me.
($2$)This is the sum of the numbers when no letters are in wrong envelop$+$exactly $2$ letters are in wrong envelops+$3$ letters+.....+$k$ letters are in wrong envelops.
Am I correct .Please inform me.
"Questions we can't answer are far better than answers we can't question"
Re: Variation Of Euler's Letter Problem
Yes, it is correct.
One one thing is neutral in the universe, that is $0$.