The 4 letters word LOVE can be arranged in 24 distinct ways. The below detailed information shows how to find how many ways are there to order the letters LOVE and how it is being calculated in the real world problems.
Distinguishable Ways to Arrange the Word LOVE
Objective: Find how many distinguishable ways are there to order the letters in the word LOVE. Step by step workout: step 1 Address the formula, input parameters and values to find how many ways are there to order the letters LOVE. Formula: nPr =n!/(n1! n2! . . . nr!) Input parameters and values: Total number of letters in LOVE: n = 4 Distinct subsets: Subsets : L = 1; O = 1; V = 1; E = 1; Subsets' count:n1(L) = 1, n2(O) = 1, n3(V) = 1, n4(E) = 1 step 2 Apply the values extracted from the word LOVE in the (nPr) permutations equation nPr = 4!/(1! 1! 1! 1! ) = 1 x 2 x 3 x 4/{(1) (1) (1) (1)} = 24/1 = 24 nPr of word LOVE = 24 Hence, The letters of the word LOVE can be arranged in 24 distinct ways.Apart from the word LOVE, 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. Answer  Hint:We can check the number of letters in the given word. Then we can consider 3 cases. Case 1 is the number of ways of arranging such that all the 4 letters are different. Then we can find the number of ways of arranging the 2 pairs of repeating letters and 1 pair of repeating letters and other 2 are different. The sum of the combinations of these will give the required number of ways. Complete step by step solution: We have the word MATHEMATICS.It has 11 letters. Out of 11, there are 8 unique and 3 of them occur twice.We need to arrange 4 letters from the word Now we can consider 3 cases,Case 1:We can consider the case where all the 4 letters are unique.It is given by the number of ways of selecting 4 letters from the unique 8 letters.$ \Rightarrow {N_1} = {}^8{P_4}$We know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ . On substituting, we get\[ \Rightarrow {N_1} = \dfrac{{8!}}{{\left( {8 - 4} \right)!}}\]On simplification, we get\[ \Rightarrow {N_1} = \dfrac{{8!}}{{4!}}\]Now we expand the factorials.\[ \Rightarrow {N_1} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!}}\]On simplification, we get\[ \Rightarrow {N_1} = 8 \times 7 \times 6 \times 5\]On multiplication, we get\[ \Rightarrow {N_1} = 1680\]Now consider case 2:We can consider the case where 2 letters are same and 2 letters are unique.There are 3 pairs of repeating letters. Out of them 2 can selected in ${}^3{C_2}$ ways. Then we can select the other two letters from any of the remaining 7 unique letters in ${}^7{C_2}$ ways. As two letters are repeating these four letters can be arranged in $\dfrac{{4!}}{{2!}}$ ways.So, number of four-letter words that can be formed in case 2 is given by their product.${N_2} = {}^3{C_1} \times {}^7{C_2} \times \dfrac{{4!}}{{2!}}$On expanding the combinations, we get$ \Rightarrow {N_2} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} \times \dfrac{{4!}}{{2!}}$On simplification, we get$ \Rightarrow {N_2} = \dfrac{{3!}}{{2!}} \times \dfrac{{7!}}{{2! \times 5!}} \times \dfrac{{4!}}{{2!}}$On expanding the factorial, we get$ \Rightarrow {N_2} = 3 \times \dfrac{{7 \times 6 \times 5!}}{{2 \times 5!}} \times 4 \times 3$On cancelling the common terms, we get$ \Rightarrow {N_2} = 3 \times \dfrac{{7 \times 6}}{2} \times 4 \times 3$On further simplification, we get$ \Rightarrow {N_2} = 756$Case 3:Now we can consider the case where the four letters are 2 pairs of repeating letters. 2 pairs can be selected in ${}^3{C_2}$ ways. As two letters are repeating twice these four letters can be arranged in $\dfrac{{4!}}{{2! \times 2!}}$ ways.So, number of four-letter words that can be formed in case 2 is given by their product.${N_3} = {}^3{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$On expanding the combinations, we get$ \Rightarrow {N_3} = \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2! \times 2!}}$On expanding the factorial, we get$ \Rightarrow {N_3} = \dfrac{{3 \times 2}}{2} \times \dfrac{{4 \times 3 \times 2}}{{2 \times 2}}$On cancelling the common terms, we get$ \Rightarrow {N_3} = 3 \times 3 \times 2$On further simplification, we get$ \Rightarrow {N_3} = 18$Now the total number of ways of arranging 4 letters is given by the sum of the number of ways in the three cases.$ \Rightarrow N = {N_1} + {N_2} + {N_3}$On substituting, we get$ \Rightarrow N = 1680 + 756 + 18$On adding, we get$ \Rightarrow N = 2454$Therefore, the number of ways in which four letters of the word MATHEMATICS can be arranged is 2454.So, the correct answer is option D. Note: We must take the repeating letters as pairs as one letter repeats only twice. We must never forget to include one of the repeating letters to the number of unique letters. We must consider all the three cases and take their sum, not the product to get the total number of ways. In the 1st case, we used the permutations instead of combinations as the order is important in arranging the letters. In the other 2 cases, after getting the possible selecting of the letters, we must multiply it with the suitable permutation in each case.
We can view the letters of the word ENGINE as a multiset with two E's, two N's, one G, and one I, that is $\{2 \cdot E, 2 \cdot I, 1 \cdot G, 1 \cdot I\}$. When we select four of these letters, we will either have four different letters, three different letters with one repeat, or two different letters with each repeated. Case 1: Four different letters. This is a permutation of the letters E, N, G, I. They can be permuted in $4!$ ways. Case 2: Three different letters with one repeated. There are two ways to select the repeated letter from $\{E, N\}$. We can choose the locations of the repeated letter in $\binom{4}{2}$ ways. There are three ways to fill the leftmost open spot with one of the other three letters and two ways to fill the last spot with one of the two letters that has not yet been selected. Hence, there are $$\binom{2}{1}\binom{4}{2}\binom{3}{1}\binom{2}{1}$$ arrangements with three different letters in which one letter is repeated. Case 3: Two different letters with both repeated. We permute two E's and two N's. There are $\binom{4}{2}$ ways to select the locations of the two E's. There is one way to fill the remaining spots with the two N's. Hence, the number of ways to arrange two different letters with both repeated is $$\binom{4}{2}$$ Thus, the total number of ways of arranging four letters of the word ENGINE is $$4! + \binom{2}{1}\binom{4}{2}\binom{3}{1}\binom{2}{1} + \binom{4}{2}$$
This section covers permutations and combinations. Arranging Objects The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1 Example How many different ways can the letters P, Q, R, S be arranged? The answer is 4! = 24. This is because there are four spaces to be filled: _, _, _, _ The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!
n! . Example In how many ways can the letters in the word: STATISTICS be arranged? There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: 10!=50 400 Rings and Roundabouts
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! Example Ten people go to a party. How many different ways can they be seated? Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440 Combinations The number of ways of selecting r objects from n unlike objects is: Example There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls? 10C3 =10!=10 × 9 × 8= 120 Permutations A permutation is an ordered arrangement.
nPr = n! . Example In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use. 10P3 =10! = 720 There are therefore 720 different ways of picking the top three goals. Probability The above facts can be used to help solve problems in probability. Example In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery? The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 . Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance. |
How many ways can a 4 letter word be arranged?
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