Question

In how many permutations of 10 things taken 4 at a time will (i) one thing always occur (ii) one thing never occurs? \[\]

Answer 1

Hint: We use the formula for combination of $r$ objects from $n$ distinct objects as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and we can arrange $r$ objects in $r!$ ways. We use the rule of product and find the number of permutations of 10 things taken 4 at a time with one thing always occurring and never occurring. \[\]

Complete step by step answer:
(i) We have to permute 4 out of 10 distinct things with the condition that 1 thing always occurs. We first select that one thing and can fix one place out of 4 places in ${}^{4}{{C}_{1}}$ ways; we can select 3 things from $10-1=9$ things in ${}^{9}{{C}_{3}}$ ways. We can arrange the things in $3!$ ways. So by rule of product number of permutations of 10 things taken 4 at a time one thing always occurs is
\[{}^{4}{{C}_{1}}\times {}^{9}{{C}_{3}}\times 3!=4\times \dfrac{9!}{3!6!}\times 3!=4\times 9\times 8\times 7=2016\]
(ii) We have to permute 4 out of 10 distinct things with the condition that 1 thing never occurs. We can exclude the one thing from our selection and we have $10-1=9$ things to select from. We can select 4 things out of 9 things in ${}^{9}{{C}_{4}}$ways and we can find then arrange them in $4!$ ways. So by rule of product number of permutations of 10 things taken 4 at a time one thing never occurs is
\[{}^{9}{{C}_{4}}\times 4!=\dfrac{9!}{4!5!}\times 4!=9\times 8\times 7\times 6=3024\]

Note: We recall that number of permutation of $n$ distinct objects in $r$ places is given by${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. We can directly solve using the formula for number of permutations of $r$ where $s$ things are always included is $r!\times {}^{n-s}{{C}_{r-s}}$ and where $s$ things are never included is ${}^{n-s}{{P}_{r}}$. We should remember the rule of product from the fundamentals of counting which states that if there are $m$ ways to do something and $n$ ways to other things then there are $m\times n$ ways to do both things.