Out of 7 consonants and 4 vowels, how many words can be formed such that it contains 3 consonants and 2 vowels ?
- 36000
- 55000
- 25200
- 75000
- None of these
Answer [Detailed Solution Below]
Option 3 : 25200
Concept:
The number of ways to select r things out of n given things wherein r ≤ n is given by: \[{\;^n}{C_r} = \frac{{n!}}{{r!\; × \left[ {n - r} \right]!}}\]
Calculation:
Given: There are 7 consonants and 4 vowels
Here, we have to find how many word can be formed such that it has 3 consonants and 2 vowels.
No. of ways to select 2 vowels out of 4 vowels = \[{\;^4}{C_2} \]
No. of ways to select 3 consonants out of 7 consonants = \[{\;^7}{C_3} \]
∴ No. of words that can be formed which contains 3 consonants and 2 vowels = \[{\;^4}{C_2} \] × \[{\;^7}{C_3} \]
As we know that, \[{\;^n}{C_r} = \frac{{n!}}{{r!\; × \left[ {n - r} \right]!}}\]
⇒ \[{\;^4}{C_2} \] × \[{\;^7}{C_3} \] = 6 × 35 = 210
The no. of ways to arrange words containing 3 consonants and 2 vowels = 210 × 5! = 25200
Hence, option C is the correct answer.
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Answer
Verified
Hint: First of all, find the number of ways of choosing the vowels and consonants separately. Then, find the ways to choose both of them. And finally, find the number of ways of arranging them by multiplying it with 5!.
Complete step-by-step answer:
In the question, we are given that out of 7 consonants and 4 vowels, we need to find how many 5 letters words consisting of 3 consonants and 2 vowels can be formed. So, first of all, we need to know what Combination in
Mathematics is. In Mathematics, the combination is a solution of items from a collection, such that the order doesn’t matter. For example, we are given 3 fruits, say an apple, an orange, and a pear. 3 combinations can be drawn from this set: an apple, a pear; an orange, a pear; an apple, an orange. More formally, a K – combination of the set S is a subset of the distinct elements of S. If a set has ‘n’ elements, the number of K – combinations is equal to binomial coefficient,
\[^{n}{{C}_{k}}=\dfrac{n\left[
n-1 \right]\left[ n-2 \right].....\left[ n-k+1 \right]}{k\left[ k-1 \right]\left[ k-2 \right].....1}\]
which can also be written using factorials as \[\dfrac{n!}{k!\left[ n-k \right]!}\] where \[k\le n\] and is 0 when k > n. The set of all k – combinations of a set often denoted by \[^{s}{{C}_{k}}\].
Now in the question, we need to choose 3 consonants and 2 vowels from the given 7 consonants and 4 vowels. The number of ways in which consonants can be chosen is
\[^{7}{{C}_{3}}=\dfrac{7!}{4!\times
3!}=35\]
Now for finding the number of ways in which 2 vowels can be selected from 4 vowels is
\[^{4}{{C}_{2}}=\dfrac{4!}{2!\times 2!}=6\]
So, the number of ways of selecting only consonants is 35 and vowels is 6. Hence, for selecting both consonants and vowels is:
\[\Rightarrow 35\times 6=210\text{ ways}\]
So now, we need to form the word by 5 distinct letters, so that can be arranged in 5! ways which is equal to 120 ways. So, the total number of words that can be formed is
\[120\times 210=25200\]
Hence, there are a total of 25200 words that can be formed.
Note: Students generally get confused between permutations and combinations as combination means only choosing while permutation means first choosing and then arranging. First we find 5 letters out of given letters and then arrange them using the formula 5! = $ 5 \times 4 \times 3 \times 2 \times 1$.
Out of 7 Consonants and 4 vowels, words are to be formed by involving 3 consonants and 2 vowels. The number of such words are formed is:
A. 25200
B. 22500
C. 10080
D. 5040
Answer
Verified
Hint- In this question we use the theory of permutation and combination. We have to choose out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed. For example, in this case, how we choose two vowels out of four vowels and this can be done in ${}^{\text{4}}{{\text{C}}_2}$ =6 ways.
Complete step-by-step answer:
As we know,
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{[n - r]}}!{\text{]}}}}$
Number
of ways of selecting [3 consonants out of 7] and [2 vowels out of 4] is calculated using the formula-
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{[n - r]}}!{\text{]}}}}$
Now,
${}^{\text{7}}{{\text{C}}_{\text{3}}} \times {}^{\text{4}}{{\text{C}}_{\text{2}}}$= $\dfrac{{{\text{7!}}}}{{{\text{[7 - 3]!3!}}}} \times \dfrac{{{\text{4!}}}}{{{\text{[4 - 2]!2!}}}}$
=
$\dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \dfrac{{4 \times 3}}{{2 \times 1}}$
= 210
Number of groups, each having 3 consonants and 2 vowels =210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves =5!
=
5×4×3×2×1
=
120
∴ Required number of ways = [210×120] =25200.
Hence, the answer is 25200.
So, option [A] is the correct answer.
Note- Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. Thus, if we want to figure out how many combinations we have of n objects then r at a time, we just create all the permutations and then divide by r! variant.