2 different vowels and 2 different consonants are to be selected from the English alphabet. Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet
\(=\space^5C_2=\frac{5!}{2!3!}=10\)
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
= \(^{21}C_2\)
= \(\frac{21!}{2!9!}\)
\(= 210\)
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words = \( 2100 \times 4!\) = 50400
\(f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\ 3x, & \quad 3≤x≤10 \end{cases}\)
The relation g is defined by
\(g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\ 3x, & \quad 2≤x≤10 \end{cases}\)
Show that f is a function and g is not a function.
Permutation is the method or the act of arranging members of a set into an order or a sequence.
Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.