Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, `O_1` and `O_2`. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, `RO_1O_2T`. Corresponding to this permutation,we have 2 ! permutations `RO_1O_2T` and `RO_2O_1T` which will be exactly the same permutation if `O_1` and `O_2` are not treated as different, i.e., if `O_1` and `O_2` are the same O at both places.
Therefore, the required number of permutations `= (4!)/(2!) = 3 xx 4 = 12`
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.
Temporarily, let us treat these letters different and name them as `I_1, I_2, T_1 , T_2, T_3`. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, `I_1 NT_1 SI_2 T_2 U E T_3`. Here if `I_1, I_2` are not same and `T_1, T_2, T_3` are not same, then `I_1, I_2` can be arranged in 2! ways and `T_1, T_2, T_3` can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation `I_1NT_1SI_2T_2UET_3`. Hence, total number of different permutations will be `(9!)/(2 ! 3!)`
We can state (without proof) the following theorems:
`color(red)("Theorem 3")` The number of permutations of n objects, where p objects are of the same kind and rest are all different `color(red)(=(n!)/(p !).)`
In fact, we have a more general theorem
`color(red)("Theorem 4")` The number of permutations of `color(blue)(n)` objects, where `color(blue)(p_1)` objects are of one kind, `color(blue)(p_2)` are of second kind, ........., `color(blue)(p_k)` are of `k^(th)` kind and the rest, if any, are of different kind is `color(red)((n!)/ (p_1! p_2! ...p_k!))`
Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, `O_1` and `O_2`. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, `RO_1O_2T`. Corresponding to this permutation,we have 2 ! permutations `RO_1O_2T` and `RO_2O_1T` which will be exactly the same permutation if `O_1` and `O_2` are not treated as different, i.e., if `O_1` and `O_2` are the same O at both places.
Therefore, the required number of permutations `= (4!)/(2!) = 3 xx 4 = 12`
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.
Temporarily, let us treat these letters different and name them as `I_1, I_2, T_1 , T_2, T_3`. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, `I_1 NT_1 SI_2 T_2 U E T_3`. Here if `I_1, I_2` are not same and `T_1, T_2, T_3` are not same, then `I_1, I_2` can be arranged in 2! ways and `T_1, T_2, T_3` can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation `I_1NT_1SI_2T_2UET_3`. Hence, total number of different permutations will be `(9!)/(2 ! 3!)`
We can state (without proof) the following theorems:
`color(red)("Theorem 3")` The number of permutations of n objects, where p objects are of the same kind and rest are all different `color(red)(=(n!)/(p !).)`
In fact, we have a more general theorem
`color(red)("Theorem 4")` The number of permutations of `color(blue)(n)` objects, where `color(blue)(p_1)` objects are of one kind, `color(blue)(p_2)` are of second kind, ........., `color(blue)(p_k)` are of `k^(th)` kind and the rest, if any, are of different kind is `color(red)((n!)/ (p_1! p_2! ...p_k!))`