Math Problem Statement

The letters of the word 'SELECTION' are to be arranged in a row. Calculate the number of arrangements if: (i) there is no restriction, (ii) it must begin and end with letter 'E', (iii) all the vowels must be separated, (iv) all the consonants are together.

Solution

Let’s address the problem step-by-step for each part of the question:

The word: SELECTION

  • It has 9 letters in total.
  • Letter distribution:
    • S, L, C, T, N → occur once each.
    • E → occurs twice.
    • I → occurs once.
    • O → occurs once.

Now, we calculate arrangements for each case.


(i) No restriction

The total number of arrangements of the 9 letters, accounting for the repetition of the letter E:

Number of arrangements=9!2!\text{Number of arrangements} = \frac{9!}{2!}

=362,8802=181,440= \frac{362,880}{2} = 181,440


(ii) It must begin and end with the letter "E"

If the arrangement must start and end with "E", these two positions are fixed. The remaining 7 letters (S, L, C, T, N, I, O) need to be arranged.

The number of arrangements is:

Number of arrangements=7!\text{Number of arrangements} = 7!

=5,040= 5,040


(iii) All the vowels must be separated

The vowels in "SELECTION" are E, E, I, and O (4 vowels). To ensure that no two vowels are adjacent, arrange the consonants (S, L, C, T, N) first, and then place the vowels in the gaps.

  1. Arrange the consonants: There are 5 consonants (S, L, C, T, N), so:

Consonant arrangements=5!\text{Consonant arrangements} = 5!

=120= 120

  1. Gaps between consonants: After arranging consonants, there are 6 gaps (e.g., _ S _ L _ C _ T _ N _). Place the vowels in these gaps, ensuring no gap has more than one vowel.
    • Select 4 gaps from 6 to place vowels: (64)=15\binom{6}{4} = 15.
    • Arrange the vowels E, E, I, O in the selected gaps, considering repetition of E:

Vowel arrangements=4!2!=12\text{Vowel arrangements} = \frac{4!}{2!} = 12

  1. Total arrangements:

Total=120×15×12=21,600\text{Total} = 120 \times 15 \times 12 = 21,600


(iv) All consonants are together

Treat all consonants (S, L, C, T, N) as a single "block." This block will be arranged along with the vowels (E, E, I, O), making a total of 5 items.

  1. Arrange the block + vowels:

Arrangements=5!2!=60\text{Arrangements} = \frac{5!}{2!} = 60

  1. Within the consonant block: The 5 consonants can be arranged among themselves:

Consonant arrangements=5!=120\text{Consonant arrangements} = 5! = 120

  1. Total arrangements:

Total=60×120=7,200\text{Total} = 60 \times 120 = 7,200


Final Results:

  1. No restriction: 181,440
  2. Begin and end with "E": 5,040
  3. All vowels separated: 21,600
  4. All consonants together: 7,200

Would you like further clarifications on any part of the solution?

Here are 5 related questions you might explore:

  1. How would the results change if there were more than two repeating letters?
  2. What if all vowels must appear together instead of being separated?
  3. How does the inclusion of more restrictions impact the number of arrangements?
  4. Can you visualize the arrangement of gaps and letters for the vowel separation case?
  5. What if one of the consonants repeats as well?

Tip: Breaking problems into smaller logical steps often simplifies complex combinatorial calculations!

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Math Problem Analysis

Mathematical Concepts

Permutations
Combinatorics
Factorial Calculations

Formulas

Total permutations with repetitions = n! / (p1! * p2! * ...)
Permutations with restrictions, such as fixed positions or gap placements

Theorems

Fundamental principle of counting

Suitable Grade Level

Grades 10-12