How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?
Given:
Total number of vowels \[=\text{ }5\]
Total number of consonants \[=\text{ }17\]
Number of ways = (No. of ways of choosing 2 vowels from 5 vowels) × (No. of ways of choosing 3 consonants from 17 consonants)
\[=\text{ }{{(}^{5}}{{C}_{2}})\text{ }\times \text{ }{{(}^{17}}{{C}_{3}})\]
By using the formula,
\[^{n}{{C}_{r}}~=\text{ }n!/r!\left( n\text{ }-\text{ }r \right)!\]
\[=\text{ }10\text{ }\times \text{ }\left( 17\times 8\times 5 \right)\]
Or,
\[=\text{ }10\text{ }\times \text{ }680\]
\[=\text{ }6800\]
Now we need to find the no. of words that can be formed by \[2\]vowels and \[3\]consonants.
The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the total no. of words that can be formed is \[5!\]
So, \[6800\text{ }\times \text{ }5!\text{ }=\text{ }6800\text{ }\times \text{ }\left( 5\times 4\times 3\times 2\times 1 \right)\]
\[=\text{ }6800\text{ }\times \text{ }120\]
\[=\text{ }816000\]
∴ The no. of words that can be formed containing \[2\]vowels and \[3\]consonants are \[816000\]
Misc 1 - Chapter 7 Class 11 Permutations and Combinations (Term 2)
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Misc 1 How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? Number ways of selecting 2 vowels & 3 consonants = 3C2 × 5C3 = 3!/2!(3 − 2)! × 5!/3!(5 − 3)! = 3!/2!1! × 5!/3!2! = 30 Now, Each of these 5 letters can be arranged in 5 ways Number of arrangements = 5P5 = 5!/(5 − 5)! = 5!/0! = 5! = 5 × 4 × 3 × 2 × 1 = 120 Thus, Total number of words = Number of ways of selecting × Number of arrangements = 30 × 120 = 3600
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How many words can be formed, each of $2$ vowels and $3$ consonants from letters of the word "DAUGHTER"
What my textbook has done: it has first taken combinations of vowels and then consonants then multiplied them altogether. Now for each combination of words they can be shuffled in $5!$ ways, so multipliying by $5!$ we get the required answer.
My question is: why has the book used combinations instead of permutations while selecting vowels and consonants?
Thanks.
JKnecht
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asked Jun 4, 2015 at 19:00
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All the letters are different, so that makes things easier.
Pick the two vowels ($_3C_2$) and pick the three consonants ($_5C_3$) and then pick what order they go in $(5!)$. So the answer is $3 \cdot 10 \cdot 120 = 3600.$
You take combinations of the vowels and consonants because the order of them doesn't matter at that point. You order them in the last step, after you've chosen which ones go in your five-letter word.
In other words, it doesn't matter that I pick $A$, then $U$, instead of $U$, then $A$. It just matters that I picked the set $(A,U)$.
answered Jun 4, 2015 at 19:12
JohnJohn
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Your query why not permutation first ? As, you have to make words of length=$5$. And of these $5$, $2$ are vowels and $3$ consonants. Since, you have to first get those $2$ vowels and $3$ consonants to make the desired word. So first operation has to be combination(selection operation), which will select $2$ vowels out of $3$ vowels(A,E,U) and then you have to select 3 consonants out of $5$(D,G,H,T,R). And they need to be multiplied, as there can be many such combinations i.e $C(3,2)*C(5,3)$. Now that you have formed $5$ letter word. These letters can be arranged among themselves to make different words. Hence, you need to apply permutation(arrangement) i.e. $5!$, making final result= $C(3,2)*C(5,3)*5!$.
answered Jun 4, 2015 at 19:40
user2016963user2016963
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