How many distinct permutations can be made from the letters of the word MATHEMATICS?

B. Compute for the number of permutations 1. How many distinct permutations can be formed from the letters of the wordMATHEMATICS?-2. How many ways can 6 different keys be arranged in a key ring?3. How many three-letter words can be formed from the letters in the wordVIRUS?4. In how many ways can 8 chairs be arranged in a row?5. In how many ways can 3 officers consisting of president, secretary, andtreasurer be chosen from a group of 11 members?​

Answer:

1. MATHEMATICS = 11 letters

P(n,r) = P(11,11)  

= 11!/(11−11)!

= 3.99168E+7

= 39,916,800

Repeated letters: 2 M, 2 A, 2 T

39,916,800/(2!2!2!) = 4,989,600 ways

2. P(n,r) = P(6,6)  

= 6!/(6−6)!

= 720 ways

3. VIRUS = 5 letters

P(n,r) = P(5,3)

= 5!/(5−3)!

= 60 ways

4. P(n,r) = P(8,8)  

= 8!/(8−8)!

= 40,320 ways

5. P(n,r) = P(11,3)  

= 11!/(11−3)!

= 990 ways

How many distinct permutations can be made from the letters of the word MATHEMATICS?

Answer:

1. 4,989,600

2. 720

3. 20

4. 42320

5. 165

Answer:

  • In 4989600 distinct ways, the letters of word "MATHEMATICS" can be arranged.

Explanation:

  • M = 2; A = 2; T = 2; H = 1; E = 1; I = 1; C = 1; S = 1;n1(M) = 2, n2(A) = 2, n3(T) = 2, n4(H) = 1, n5(E) = 1, n6(I) = 1, n7(C) = 1, n8(S) = 1

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

Answer

Verified

Hint:Here, we will proceed by observing all the letters in the word MATHEMATICS that are repeating and then, we will use the formula i.e., Permutation of n items out of which x items, y items and z items of different types are repeating = $\dfrac{{n!}}{{x!y!z!}}$. For the next two parts, we will fix the first letter of the word as C and T in order to find out the different arrangements possible.Complete step-by-step answer:
The word MATHEMATICS consists of 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I, 1 C and 1 S.
Total number of letters in the word MATHEMATICS = 11
As we know that
Total number of different arrangements of n items out of which x items, y items and z items of different types are repeating = $\dfrac{{n!}}{{x!y!z!}}{\text{ }} \to {\text{(1)}}$
Using the formula given by equation (1), we can write
Total number of different arrangements which can be made by using all the 11 letters in the word MATHEMATICS in which letter M, letter A and letter T are repeating twice = $\dfrac{{11!}}{{2!2!2!}} = \dfrac{{11.10.9.8.7.6.5.4.3.2!}}{{2.1.2.1.2!}} = \dfrac{{11.10.9.8.7.6.5.4.3}}{4} = 4989600$
Therefore, a total of 4989600 words can be formed using all the letters of the word MATHEMATICS.

For the words which begin with letter C formed using all the letters of the word MATHEMATICS, the first letter is fixed as C so the next 10 letters need to be selected from the left letters (i.e., 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I and 1 S)
Using the formula given by equation (1), we can write
Total number of different arrangements which can be made by using all the left 10 letters (except letter C) in the word MATHEMATICS in which letter M, letter A and letter T are repeating twice = $\dfrac{{10!}}{{2!2!2!}} = \dfrac{{10.9.8.7.6.5.4.3.2!}}{{2.1.2.1.2!}} = \dfrac{{10.9.8.7.6.5.4.3}}{4} = 453600$
Therefore, a total of 453600 words which begin with C can be formed using all the letters of the word MATHEMATICS.

For the words which begin with letter T formed using all the letters of the word MATHEMATICS, the first letter is fixed as T so the next 10 letters need to be selected from the left letters (i.e., 2 M’s, 2 A’s, 1 T, 1 H, 1 E, 1 I, 1 C and 1 S)
Also we know that
Total number of different arrangements of n items out of which x items and y items of different types are repeating = $\dfrac{{n!}}{{x!y!}}{\text{ }} \to {\text{(2)}}$
Using the formula given by equation (2), we can write
Total number of different arrangements which can be made by using all the left 10 letters (except one of the two letters T) in the word MATHEMATICS in which letter M, letter A are repeating twice = $\dfrac{{10!}}{{2!2!}} = \dfrac{{10.9.8.7.6.5.4.3.2!}}{{2.1.2!}} = \dfrac{{10.9.8.7.6.5.4.3}}{2} = 907200$
Therefore, a total of 907200 words which begin with T can be formed using all the letters of the word MATHEMATICS.

Note- In this particular problem, since we have to rearrange the letters of the word MATHEMATICS that’ s why we are using permutation formulas. If we were asked for selection of some letters out of all the letters we would have used combinations formula. The general formula for arrangement of r items out of n items is given by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\].

How many four letter permutations can be formed from the letters of the word MATHEMATICS?

Therefore, the number of ways in which four letters of the word MATHEMATICS can be arranged is 2454.

How many letters are there in the word MATHEMATICS?

Solution : (i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different.