Answer :

In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R.

Number of ways of selecting 2 vowels out of 3 vowels $={ }^{3} C_2=3$

Number of ways of selecting 3 consonants out of 5 consonants $={ }^{5} C_3=10$

Therefore, number of combinations of 2 vowels and 3 consonants $=3 \times 10=30$

Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5 ! ways.

Hence, required number of different words $=30 \times 5 !=3600$

Answer :

In the word EQUATION, there are 5 vowels, namely, $A, E, I, O$, and $U$, and 3 consonants, namely, Q, $T$, and $N$. Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be ${ }^{2} P_2=2$ ! Corresponding to each of these permutations, there are 5 ! permutations of the five vowels taken all at a time and 3 ! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words $=2 ! \times 5 ! \times 3 !$

$=1440$

Answer :

A committee of 7 has to be formed from 9 boys and 4 girls.

i. Since exactly 3 girls are to be there in every committee, each committee must consist of $(7-3)$ boys only.

Thus, in this case, required number of ways $=$

$ { }^{4} C_3 \times{ }^{9} C_4=\dfrac{4 !}{3 ! 1 !} \times \dfrac{9 !}{4 ! 5 !} $

$=4 \times \dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}$

$=504$

(ii) Since at least 3 girls are to be there in every committee, the committee can consist of

Answer :

In the given word EXAMINATION, there are 11 letters out of which, $A, I$, and $N$ appear 2 times and all the other letters appear only once.

The words that will be listed before the words starting with $E$ in a dictionary will be the words that start with $A$ only.

Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.

Since there are 2 Is and $2 Ns$ in the remaining 10 letters,

Number of words starting with $A=\dfrac{10 !}{2 ! 2 !}=907200$

Thus, the required numbers of words is 907200 .

Answer :

A number is divisible by 10 if its units digits is 0 .

Therefore, 0 is fixed at the units place.

Therefore, there will be as many ways as there are ways of filling 5 vacant places

$ \begin{array}{|c|c|c|c|c|c|} \hline & & & & & 0\\ \hline \end{array} $

by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9 ).

The 5 vacant places can be filled in 5 ! ways.

Hence, required number of 6-digit numbers $=5 !=120$

Answer :

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=\dfrac{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=\dfrac{21 !}{2 ! 19 !}=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 $=2100 \times 4 !=50400$

Answer :

It is given that the question paper consists of 12 questions divided into two parts Part I and Part II, containing 5 and 7 questions, respectively.

A student has to attempt 8 questions, selecting at least 3 from each part.

This can be done as follows.

Answer :

From a deck of 52 cards, 5 -card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.

In a deck of 52 cards, there are 4 kings.

1 king can be selected out of 4 kings in ${ }^{4} C_1$ ways.

4 cards out of the remaining 48 cards can be selected in ${ }^{48} C_4$ ways.

Thus, the required number of 5-card combinations is ${ }^{4} C_1 \times{ }^{48} C_4$.

Answer :

5 men and 4 women are to be seated in a row such that the women occupy the even places.

The 5 men can be seated in 5 ! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).

$ M \times M \times M \times M \times M $

Therefore, the women can be seated in 4! ways.

Thus, possible number of arrangements $=4 ! \times 5 !=24 \times 120=2880$

Answer :

From the class of 25 students, 10 are to be chosen for an excursion party.

Since there are 3 students who decide that either all of them will join or none of them will join, there are two cases.

Case I: All the three students join.

Then, the remaining 7 students can be chosen from the remaining 22 students in ${ }^{22} C_7$ ways.

Case II: None of the three students join.

Then, 10 students can be chosen from the remaining 22 students in ${ }^{22} C _{10}$ ways.

Thus, required number of ways of choosing the excursion party is ${ }^{22} C_7+{ }^{22} C _{10}$.

Answer :

In the given word ASSASSINATION, the letter A appears 3 times, $S$ appears 4 times, I appears 2 times, $N$ appears 2 times, and all the other letters appear only once.

Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects.

These 10 objects in which there are 3 As, 2 ls, and 2 Ns can be arranged in $\dfrac{10 !}{3 ! 2 ! 2 !}$ ways.

Thus, required number of ways of arranging the letters of the given word

$=\dfrac{10 !}{3 ! 2 ! 2 !}=151200$