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Question: How many words can be formed from the letters of the word TRIANGLE? In how many of these does the wo...

How many words can be formed from the letters of the word TRIANGLE? In how many of these does the word start with T and ends with E?

Explanation

Solution

Here we calculate the number of letters in the given word. We calculate ways to fill each position using the concept of combination where we reduce the number of letters available for the next position by one after filling each position.

  • Combination is given by nCr=n!(nr)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}, where n is the total number of available objects and r is the number of objects we have to choose.
  • Factorial terms open up as n!=n(n1)!n! = n(n - 1)!

Complete step-by-step answer:
We are given the word TRIANGLE. We count the number of letters is 8.
So we have to count the number of 8 letter words that can be formed from the letters of the word TRIANGLE.
We have to fill 8 positions with 8 letters. Since each position can have one letter at a time so we keep the value of r as 1.
We use the method of combination nCr=n!(nr)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
Whenever the value of r is 1 we can write nC1=n!(n1)!1!^n{C_1} = \dfrac{{n!}}{{(n - 1)!1!}}
From the formula of factorial we know n!=n(n1)!n! = n(n - 1)!
So we can write nC1=n(n1)!(n1)!^n{C_1} = \dfrac{{n(n - 1)!}}{{(n - 1)!}}
On cancelling same terms from numerator and denominator we get
nC1=n^n{C_1} = n … (1)
First Position:
We have total number of letters to choose from as 8, so value of n=8n = 8
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill first position is 8C1^8{C_1}
Using the equation (1) we get 8C1=8^8{C_1} = 8
So, number of ways to fill first position is 8.
Second Position:
We have total number of letters to choose from as 7, so value of n=7n = 7
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill second position is 7C1^7{C_1}
Using the equation (1) we get 7C1=7^7{C_1} = 7
So, number of ways to fill second position is 7.
Third Position:
We have total number of letters to choose from as 6, so value of n=6n = 6
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill third position is 6C1^6{C_1}
Using the equation (1) we get 6C1=6^6{C_1} = 6
So, number of ways to fill third position is 6.
Fourth Position:
We have total number of letters to choose from as 5, so value of n=5n = 5
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill fourth position is 5C1^5{C_1}
Using the equation (1) we get 5C1=5^5{C_1} = 5
So, number of ways to fill fourth position is 5.
Fifth Position:
We have total number of letters to choose from as 4, so value of n=4n = 4
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill fifth position is 4C1^4{C_1}
Using the equation (1) we get 4C1=4^4{C_1} = 4
So, number of ways to fill fifth position is 4.
Sixth Position:
We have total number of letters to choose from as 3, so value of n=3n = 3
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill sixth position is 3C1^3{C_1}
Using the equation (1) we get 3C1=3^3{C_1} = 3
So, number of ways to fill sixth position is 3.
Seventh Position:
We have total number of letters to choose from as 2, so value of n=2n = 2
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill seventh position is 2C1^2{C_1}
Using the equation (1) we get 2C1=2^2{C_1} = 2
So, number of ways to fill seventh position is 2.
Eighth Position:
We have total number of letters to choose from as 1, so value of n=1n = 1
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill eighth position is 1C1^1{C_1}
Using the equation (1) we get 1C1=1^1{C_1} = 1
So, number of ways to fill eighth position is 1.
Number of total words formed from the letters of the word TRIANGLE is given by multiplication of number of ways to fill each position.
Therefore, total number of words formed =8×7×6×5×4×3×2×1 = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
On multiplying the values we get number of total words =40,320 = 40,320
Now we have to find the words that start with T and end with E.
So we fix the letters for the first and the eighth position and find number of ways to fill all the remaining positions.
So, number of ways to fill first position is 1 (only T)
Number of ways to fill eighth position is 1(only E)
Now we are left with 6 letters,
Number of ways to fill second position will be 6
Number of ways to fill third position will be 5
Number of ways to fill fourth position will be 4
Number of ways to fill fifth position will be 3
Number of ways to fill sixth position will be 2
Number of ways to fill seventh position will be 1
Using the same concept as used for total number of word formed, we can write number of words formed will be 1×6×5×4×3×2×1×11 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1
Calculating the multiplication we get value =720 = 720
Therefore, the total number of words formed from the word TRIANGLE is 40,320 out of which 720 words start with T and end with E.

Note: Alternative method:
Number of ways to fill each position is given by n!n!where n is the number of letters available.
So, number of 8 letter words formed is given by 8!8!
Use the expansion of factorial i.e. n!=n.(n1).(n2)......3.2.1n! = n.(n - 1).(n - 2)......3.2.1
8!=8×7×6×5×4×3×2×1\Rightarrow 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
8!=40,320\Rightarrow 8! = 40,320
So, total number of words formed is 40,320
Now we have to find the words that start with T and end with E.
So we fix the letters for the first and the eighth position and find a number of ways to fill all the remaining positions.
So, number of ways to fill first position is 1 (only T)
Number of ways to fill eighth position is 1(only E)
Now to fill 6 positions we have 6 letters.
Number of words formed is 1×6!×11 \times 6! \times 1
Use the expansion of factorial i.e. n!=n.(n1).(n2)......3.2.1n! = n.(n - 1).(n - 2)......3.2.1
6!=6×5×4×3×2×1\Rightarrow 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1
6!=720\Rightarrow 6! = 720
So, the number of words formed is 720.