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Question: In how many permutations of 10 things taken 4 at a time will (i) one thing always occur (ii) one thi...

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

Explanation

Solution

We use the formula for combination of rr objects from nn distinct objects as nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} and we can arrange rr objects in r!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 4C1{}^{4}{{C}_{1}} ways; we can select 3 things from 10−1=910-1=9 things in 9C3{}^{9}{{C}_{3}} ways. We can arrange the things in 3!3! ways. So by rule of product number of permutations of 10 things taken 4 at a time one thing always occurs is
4C1×9C3×3!=4×9!3!6!×3!=4×9×8×7=2016{}^{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=910-1=9 things to select from. We can select 4 things out of 9 things in 9C4{}^{9}{{C}_{4}}ways and we can find then arrange them in 4!4! ways. So by rule of product number of permutations of 10 things taken 4 at a time one thing never occurs is
9C4×4!=9!4!5!×4!=9×8×7×6=3024{}^{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 nn distinct objects in rr places is given bynPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. We can directly solve using the formula for number of permutations of rr where ss things are always included is r!×n−sCr−sr!\times {}^{n-s}{{C}_{r-s}} and where ss things are never included is n−sPr{}^{n-s}{{P}_{r}}. We should remember the rule of product from the fundamentals of counting which states that if there are mm ways to do something and nn ways to other things then there are m×nm\times n ways to do both things.