The English alphabet has $5$ vowels and $21$ consonants. How many words with two different vowels and $2$ different consonants can be formed from the alphabet?
$2$ 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 $=\,^{5} C_{2}=\frac{5 !}{2 ! 3 !}=10$
since there are $21$ consonants in the English alphabet, number of ways of selecting $2$ different consonants from the alphabet $=\,^{21} C_{2}=\frac{21 !}{2119 !}=210$
Therefore, number of combinations of $2$ different vowels and $2$ different consonants $=10 \times 210=2100$
Each of these $2100 $ combinations has $4$ letters, which can be arranged among themselves in $4 !$ ways.
Therefore, required number of words $=2100 \times 4 !=50400$
If $^{n}{P_4} = 24.{\,^n}{C_5},$ then the value of $n$ is
The least value of natural number $n$ satisfying $C(n,\,5) + C(n,\,6)\,\, > C(n + 1,\,5)$ is
The number of ways in which thirty five apples can be distributed among $3$ boys so that each can have any number of apples, is
Let $A = \left\{ {{a_1},\,{a_2},\,{a_3}.....} \right\}$ be a set containing $n$ elements. Two subsets $P$ and $Q$ of it is formed independently. The number of ways in which subsets can be formed such that $(P-Q)$ contains exactly $2$ elements, is
If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2+......+ ^{2017}C_{1008} = \lambda ^2 (\lambda > 0),$ then remainder when $\lambda $ is divided by $33$ is-