Question

Find the number of arrangements that can be made out of the letters of the words (1) independence, (2) superstitious, (3) institutions

Answer 1

Hint: This question can be solved using the concept of permutation and to find the required answer we will apply the formula for permutation of n items out of which some items are repeating different times also remember to use formula $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times .....1$.

Complete step-by-step answer:
(1) independence word contains 12 letters in which letter n is repeating 3 times, d is repeating 2 times, e is repeating 4 times.
According to permutations, we know that if a word of n letters in which three different letters are repeating x, y and z times respectively then the total number of words which can be obtained by rearranging the letters in the given word are$\dfrac{{n!}}{{x!y!z!}}$.
So, here ${\text{n}} = 12,{\text{ x}} = 3,{\text{ y}} = 2$ and ${\text{z}} = 4$
Therefore, total number of arrangements made by rearranging the letters in the word independence$ = \dfrac{{12!}}{{3!2!4!}} = \dfrac{{12.11.10.9.8.7.6.5.4!}}{{3.2.1.2.1.4!}} = \dfrac{{12.11.10.9.8.7.6.5}}{{3.2.1.2.1}} = 1663200$
Hence, the total number of arrangements made by rearranging the letters in the word independence are 1663200.

(2) superstitious words contain 13 letters in which letter s is repeating 3 times, u is repeating 2 times, t is repeating 2 times, i is repeating 2 times.
According to permutations, we know that if a word of n letters in which four different letters are repeating x, y, z and w times respectively then the total number of words which can be obtained by rearranging the letters in the given word are$\dfrac{{{\text{n!}}}}{{{\text{x!y!z!w!}}}}$.
So, here ${\text{n}} = 13,{\text{ }}x = 3,{\text{ y}} = 2,{\text{ z}} = {\text{2}}$ and ${\text{w}} = 2$
Therefore, total number of arrangements made by rearranging the letters in the word superstitious${\text{ = }}\dfrac{{13!}}{{3!2!2!2!}} = \dfrac{{13.12.11.10.9.8.7.6.5.4.3!}}{{2.1.2.1.2.1.3!}} = \dfrac{{13.12.11.10.9.8.7.6.5.4}}{{2.1.2.1.2.1}} = 129729600$
Hence, the total number of arrangements made by rearranging the letters in the word superstitious are 129729600.

(3) institutions word contains 12 letters in which letter i is repeating 3 times, n is repeating 2 times, s is repeating 2 times, t is repeating 3 times.
According to permutations, we know that if a word of n letters in which four different letters are repeating x, y, z and w times respectively then the total number of words which can be obtained by rearranging the letters in the given word are$\dfrac{{{\text{n!}}}}{{{\text{x!y!z!w!}}}}$.
So, here $x = 3,{\text{ y}} = 2,{\text{ z}} = {\text{2}}$ and ${\text{w}} = 3$
Therefore, total number of arrangements made by rearranging the letters in the word institutions${\text{ = }}\dfrac{{12!}}{{3!2!2!3!}} = \dfrac{{12.11.10.9.8.7.6.5.4.3!}}{{3.2.1.2.1.2.1.3!}} = \dfrac{{12.11.10.9.8.7.6.5.4}}{{3.2.1.2.1.2.1}} = 3326400$
Hence, the total number of arrangements made by rearranging the letters in the word institutions are 3326400.

Note: These types of problems can be solved by simply evaluating the total number of letters present in the given word and how many times different letters are repeating in the given word. Then, finally applying the formula for permutations with repeating to obtain the total number of rearrangements which can be done with the letters present in the given word.