How many ways can the letters of the word permutations be arranged if the I vowels are all together II there are always 4 letters between P and S?
In how many ways can the letters of the word PERMUTATIONS be arranged so that:i) Words starting with P and ending with S,ii) Vowels are all together,iii) There are always 4 letters between P and S? Show
Answer Verified
Hint: Here, we will use the Permutation formula which states that a permutation of any set is known as the arrangement of its members or items into linear order or if that particular set is already ordered rather than a re-arrangement of its elements. For example, we have a set of three numbers, i.e. \[\left\{ {1,2,3} \right\}\] than there will be six ways to rearrange the elements inside it as shown below: Complete step-by-step solution:
Now, for filling the remaining \[10\] positions we will use permutation property, so the arrangement will be: When the letter starts with P and ends with S= \[\dfrac{{10!}}{{2!}}\] (\[\because \]letter T is repeating twice) So, the number of arrangements will be =\[\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}\] (for example\[3! = 3 \times 2 \times 1\]) \[ \Rightarrow \]The number of arrangements (when word starts with the letter P and ends with S) =\[1814400\] ways. Step 2: For finding the arrangements when vowels are together in the word PERMUTATIONS, first we will check how many numbers of vowels are there: There are \[5\] vowels (E, A, U, I, O) in the word PERMUTATIONS and \[7\] consonants. Now, for calculating the number of ways first we will consider all the five letters of the vowels as a single letter. Now except vowels, there are \[7\] letters and one is for vowels so there will be total \[8\]letters to be arranged. Now we can arrange the \[8\] letters as= \[8!\] Also, vowels can interchange their positions in five number of ways = \[5!\] \[ \Rightarrow \]Total number of ways = \[8! \times \dfrac{{5!}}{{2!}}\] (we are dividing by \[2!\] because T comes two times in the word) \[ \Rightarrow \]Total number of ways= \[8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}\] By dividing \[\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}\], we get: \[ \Rightarrow \]Total number of ways= \[8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 5 \times 4 \times 3\] By doing simple multiplication of the term \[8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 5 \times 4 \times 3\], we get: \[ \Rightarrow \]Total number of ways (when vowels come together) =\[2419200\] Step 3: For finding the number of arrangements in the letter PERMUTATIONS when there will exactly four letters come in between the letters P and S, first we will check that in how many ways we can fill the positions of letters as shown below: $1^{st}$ way- When the letter P comes at the very first position from starting: P_ _ _ _ S _ _ _ _ _ _ $2^{nd}$ way- When the letter P comes at 2nd position: _ P_ _ _ _ S _ _ _ _ _ $3^{rd}$ way- When the letter P comes at 3rd position: _ _ P _ _ _ _ S _ _ _ _ $4^{th}$ way- When the letter P comes at 4th position: _ _ _ P _ _ _ _ S _ _ _ $5^{th}$ way: When the letter P comes at 5th position: _ _ _ _ P _ _ _ _ S _ _ $6^{th}$ way: When the letter P comes at 6th position: _ _ _ _ _ P _ _ _ _ S _ $7^{th}$ way: When the letter P comes at 7th position: _ _ _ _ _ _ P _ _ _ _ S So, there are total \[7\] ways for filling the spaces between P and S having \[4\] letters between them. Now, letters P and S can also change their position in the same way. So, again there will be \[7\] ways for filling the spaces between S and P having \[4\] letters between them. Hence, P and S or S and P can be filled in \[7 + 7 = 14\] many ways. Also, there are total \[12\] spaces which need to be filled and two positions are filled by P and S so remaining positions will be \[10\] and the number of ways for filling these positions will be: \[ \Rightarrow \]Number of ways for filling \[10\] positions= \[\dfrac{{10!}}{{2!}}\] ( we are dividing by \[2!\] because T comes two times in the word) So, the total number of ways will be equals to the product of the number of ways letter P and S can be filled with the number of ways for filling the remaining \[10\] positions: \[ \Rightarrow \]Total Number of ways (exactly \[4\]letters are coming in between P and S) = \[14 \times \dfrac{{10!}}{{2!}}\] By solving the term \[\dfrac{{10!}}{{2!}}\], we get: \[ \Rightarrow \]Total Number of ways (exactly \[4\]letters are coming in between P and S) = \[14 \times \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}\] By dividing the term \[\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}\] we get: \[ \Rightarrow \]Total Number of ways (exactly \[4\]letters are coming in between P and S) = \[14 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3\] By doing multiplication of the term \[14 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3\], we get: \[ \Rightarrow \]Total Number of ways (exactly \[4\] letters are coming in between P and S) = \[25401600\] (i) Number of arrangements (when word starts with letter P and ends with S) =\[1814400\] ways. Note:
In these types of permutation questions students generally get confused in calculating the number of ways for a word when any letter is repeating itself twice or thrice. For example, in a word KAJAL, letter A is repeating itself twice so at the time of finding the number of arrangements of letters we need to divide the result with the repeating letter case as shown below: How many ways can the letters of the word permutations be arranged if the I vowels are all together ii words start with P and ends with S?Thus total number of permutations = 14×1814400=25401600.
How many ways can the letters of the word permutations be arranged if the i words start with P and end with S II there are always 4 letters between P and S?There are in total 7 ways in which there are 4 letters between P and S. If we interchange P and S in the above table, we get 7 more ways of placing P and S. Thus, total number of ways in which P and S can be placed = 7 + 7 = 14.
...
. How many arrangements can be made with the letters of the word mathematics if I vowels occur together II consonants occur together?∴ Required number of words = (10080 x 12) = 120960.
How many 4 letter words can be formed using the letter of the word permutations?Solution. ∴ 840 four-letter words can be formed when the repetition of letters is not allowed. Report Error Is there an error in this question or solution?
|