Distinguishable Ways to Arrange the Word MATHEMATICS The below step by step work generated by the word permutations calculator shows how to find how many different ways can the letters of the word MATHEMATICS be arranged.
Objective: Find how many distinguishable ways are there to order the letters in the word MATHEMATICS.Step by step workout: step 1 Address the formula, input parameters and values to find how many ways are there to order the letters MATHEMATICS. Formula: nPr =n!/(n1! n2! . . . nr!)Input parameters and values: Total number of letters in MATHEMATICS: n = 11
Distinct subsets: Subsets : M = 2; A = 2; T = 2; H = 1; E = 1; I = 1; C = 1; S = 1; Subsets' count: n1(M) = 2, n2(A) = 2, n3(T) = 2,
n4(H) = 1, n5(E) = 1, n6(I) = 1, n7(C) = 1, n8(S) = 1
step 2 Apply the values extracted from the word MATHEMATICS in the (nPr) permutations equation nPr = 11!/(2! 2! 2! 1! 1! 1! 1! 1! )
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11/{(1 x 2) (1 x 2) (1 x 2) (1) (1) (1) (1) (1)}
= 39916800/8
= 4989600 nPr of word
MATHEMATICS = 4989600
Hence, The letters of the word MATHEMATICS can be arranged in 4989600 distinct ways.
Apart from the word MATHEMATICS, you may try different words with various lengths with or without repetition of letters to observe how it affects the nPr word permutation calculation to find how many ways the letters in the given word can be arranged.
Explanation:
For problems, like these, we need to consider the number of total letters and the number of repeated letters.
There are 9 letters in this word, so if all the letters were different there would be #9!# ways of arranging them.
However, we have #3# R's, #2# A's and #2# E's.
So, this expression becomes #(9!)/(3! xx 2! xx 2!) = (362,280)/(6 xx 2
xx 2) = (362,280)/24 = 15,120#.
Hopefully this helps!
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Find the number of different arrangements of the eleven letters in the word "PERSONALITY" if the arrangements are such that S, O and N are separated.
My Solution:
Find total number of ways to arrange 11 letters.
Find number of ways by
having S,O and N together.
Use 1 - 2.
$11! - (9 \cdot 8! \cdot 3!)= 37 739 520$
But correct answer is $8! \cdot 9C3 \cdot 3! = 20 321 280$.
May I know what have I done wrong?
asked Feb 28 at 16:19
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I suppose what the problem means by "S, O and N are separated" is that no two of them are adjacent. You've only subtracted out the ones that have these three letters collectively forming a single block. You've
yet to exclude the ones with S-O adjacent, S-N adjacent and O-N adjacent.
The answer $8! \cdot 9C3 \cdot 3!$ suggests that such an arrangement that satisfies the requirements can be achieved by first arranging the 8 letters other than S, O and N (there are 8! ways to do so). Then, there are 9 places (the leftmost and rightmost positions, plus 7 gaps in between the 8 letters) among which you choose 3 to place S, O and N. Note that you can only place one of S, O and N in one
gap, otherwise you will have two of them adjacent and violating the requirements. After choosing the 3 positions, you can then arrange S, O and N in $3!$ ways. The result then follows by the multiplication principle.
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