
Find the number of permutations of $n$distinct things taken $r$together, in which $3$particular things must occur together.
A. $^{n - 3}{C_{r - 3}} \times 3! \times (r - 2)!$
B. \[^{n - 2}{C_{r - 2}}{ \times ^{r - \,}}^2{C_1} \times 3! \times (r - 2)!\]
C. $^{n - 3}{C_{r - 3}}{ \times ^{r - \,}}^3{C_1} \times 3! \times (r - 3)!$
D. \[^{n - 2}{C_{r - 3}}{ \times ^{r - \,}}^2{C_1} \times 3! \times (r - 2)!\]
Answer
597.9k+ views
Hint: We are given the number of permutations of $n$distinct things are taken $r$together in which $3$ particular things must occur together.
Complete step by step solution:
The number of combination of $n$ distinct things $r$ together $ = {\,^n}{C_r}$
Now 3 things occur together
So, total available numbers $ = n - 3$
Total numbers $ = r - 3$
The number $ = {\,^{n - 3}}{C_{r - 3}}$
$3$Things can be arranged as $(r - 2)!$ ways and these $3$ things can be placed in $3!$ ways.
Thus, required number of things $ = {\,^{n - 3}}{C_{r - 3}}(r - 2)!3!$
Hence, the correct option is A.
Note: Students must consider $3$ combinations which are occurring and students should not forget to consider all the three cases before getting the answer.
Complete step by step solution:
The number of combination of $n$ distinct things $r$ together $ = {\,^n}{C_r}$
Now 3 things occur together
So, total available numbers $ = n - 3$
Total numbers $ = r - 3$
The number $ = {\,^{n - 3}}{C_{r - 3}}$
$3$Things can be arranged as $(r - 2)!$ ways and these $3$ things can be placed in $3!$ ways.
Thus, required number of things $ = {\,^{n - 3}}{C_{r - 3}}(r - 2)!3!$
Hence, the correct option is A.
Note: Students must consider $3$ combinations which are occurring and students should not forget to consider all the three cases before getting the answer.
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