Number of Combinations Without Repetition
Category : JEE Main & Advanced
The number of combinations (selections or groups) that can be formed from \[n\] different objects taken \[r(0\le r\le n)\] at a time is \[^{n}{{C}_{r}}=\frac{n\,!}{r\,!(n-r)\,!}\]. Also \[^{n}{{C}_{r}}={{\,}^{n}}{{C}_{n-r}}\].
Let the total number of selections (or groups) \[=x\]. Each group contains \[r\] objects, which can be arranged in \[r\,\,!\] ways. Hence the number of arrangements of \[r\] objects \[=x\times \,(r!)\]. But the number of arrangements \[=\,{{\,}^{n}}{{P}_{r}}\].
\[\Rightarrow \,x(r!)={{\,}^{n}}{{P}_{r}}\Rightarrow x=\frac{^{n}{{P}_{r}}}{r\,!}\Rightarrow x=\frac{n!}{r\,!\,(n-r)\,!}{{=}^{n}}{{C}_{r}}\].
You need to login to perform this action.
You will be redirected in
3 sec
