**Q.11. **In how many ways can the letters of the word ASSASSINATION be arranged

so that all the S’s are together?

In the word ASSASSINATION, we have
number of S =4
number of A =3
number of I= 2
number of N =2
Rest of letters appear at once.
Since all words have to be arranged in such a way that all the S are together so we can assume SSSS as an object.
The single object SSSS with other 9 objects is counted as 10.
These 10 objects can be arranged in (we have 3 A's,2 I's,2 N's)
...

**Q.10. **From a class of 25 students, 10 are to be chosen for an excursion party. There

are 3 students who decide that either all of them will join or none of them will

join. In how many ways can the excursion party be chosen ?

From a class of 25 students, 10 are to be chosen for an excursion party.
There are 3 students who decide that either all of them will join or none of them will join, there are two cases :
The case I: All 3 off them join.
Then, the remaining 7 students can be chosen from 22 students in ways.
Case II : All 3 of them do not join.
Then,10 students can be chosen from 22 students in ways.
Thus,...

**Q.9. **It is required to seat 5 men and 4 women in a row so that the women occupy the

even places. How many such arrangements are possible?

It is required to seat 5 men and 4 women in a row so that the women occupy the even places.
The 5 men can be seated in ways.
4 women can be seated at cross marked places (so that women occupy even places)
Therefore, women can be seated in ways.
Thus, the possible arrangements

**Q.8. **Determine the number of 5-card combinations out of a deck of 52 cards if each

selection of 5 cards has exactly one king.

From a deck of 52 cards, 5 cards combinations have to be made in such a way that in each selection of 5 cards there is exactly 1 king.
Number of kings =4
Number of ways of selecting 1 king
4 cards from the remaining 48 cards are selected in ways.
Thus, the required number of 5 card combinations

**Q.7.** In an examination, a question paper consists of 12 questions divided into two

parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student

is required to attempt 8 questions in all, selecting at least 3 from each part. In

how many ways can a student select the questions?

It is given that a question paper consists of 12 questions divided in two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively.
A student is required to attempt 8 questions in all, selecting at least 3 from each part.
This can be done as follows:
(i) 3 questions from part I and 5 questions from part II
(ii) 4 questions from part I and 4 questions from part II
(iii)...

**Q.6 **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?

Two different vowels and 2 different consonants are to be selected from the English alphabets.
Since there are 5 different vowels so the number of ways of selecting two different vowels =
Since there are 21 different consonants so the number of ways of...

**Q.5. **How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9

which are divisible by 10 and no digit is repeated?

For the number to be divisible by 10, unit digit should be 0.
Thus, 0 is fixed at a unit place.
Therefore, the remaining 5 places should be filled with 1,3,5,7,9.
The remaining 5 vacant places can be filled in ways.
Hence, the required number of 6 digit numbers which are divisible by 10

**Q.4. **If the different permutations of all the letter of the word EXAMINATION are

listed as in a dictionary, how many words are there in this list before the first

word starting with E?

In the word EXAMINATION, we have 11 letters out of which A,I, N appear twice and all other letters appear once.
The word that will be listed before the first word starting with E will be words starting with A.
Therefore, to get the number of words starting with A, letter A is fixed at extreme left position, the remaining 10 letters can be arranged.
Since there are 2 I's and 2 N's in the...

**Q.3 **A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways

can this be done when the committee consists of:

** (iii)** atmost 3 girls?

There are 9 boys and 4 girls. A committee of 7 has to be formed.
(ii) atmost 3 girls, there can be 4 cases :
(a) Girls =0, so boys in committee= 7-0=7
Thus, the required number of ways
(b) Girls =1, so boys in committee= 7-1=6
Thus, the required number of ways
...

**Q.3. **A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways

can this be done when the committee consists of:

**(ii) at least** 3 girls?

There are 9 boys and 4 girls. A committee of 7 has to be formed.
(ii) at least 3 girls, there can be two cases :
(a) Girls =3, so boys in committee= 7-3=4
Thus, the required number of ways
(b) Girls =4, so boys in committee= 7-4=3
Thus, the...

**Q.3. **A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways

can this be done when the committee consists of:

** (i) **exactly 3 girls?

There are 9 boys and 4 girls. A committee of 7 has to be formed.
Given : Girls =3, so boys in committee= 7-3=4
Thus, the required number of ways

**Q.2. **How many words, with or without meaning, can be formed using all the letters

of the word EQUATION at a time so that the vowels and consonants occur together?

In the word EQUATION, we have
vowels = 5(A,E,I,O,U)
consonants = 3(Q,T,N)
Since all the vowels and consonants occur together so (AEIOU) and (QTN) can be assumed as single objects.
Then, permutations of these two objects taken at a time
Corresponding to each of these permutations, there are permutations for vowels and permutations for consonants.
Thus, by multiplication principle, required...

**Q.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?

In the word DAUGHTER, we have
vowels = 3(A,E,U)
consonants = 5(D,G,H,T,R)
Number of ways of selecting 2 vowels
Number of ways of selecting 3 consonants
Therefore, the number of ways of selecting 2 vowels and 3 consonants
Each of these 30 combinations of 2 vowels and 3 consonants can...

**Q.9.** In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

9 courses are available and 2 specific courses are compulsory for every student.
Therefore, every student has to select 3 courses out of the remaining 7 courses.
This can be selected in ways.
Thus, using multiplication priciple, number of ways of selecting courses
...

**Q.8. ** A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

A bag contains 5 black and 6 red balls.
2 black balls can be selected in ways and 3 red balls can be selected in ways.
Thus, using multiplication priciple, number of ways of selecting 2 black and 3 red balls
...

**Q.7.** In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Out off, 17 players, 5 are bowlers.
A cricket team of 11 is to be selected such that there are exactly 4 bowlers.
4 bowlers can be selected in ways and 7 players can be selected in ways.
Thus, using multiplication priciple, number of ways of selecting the team
...

**Q.6. ** Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

In a deck, there is 4 ace out of 52 cards.
A combination of 5 cards is to be selected containing exactly one ace.
Then, one ace can be selected in ways and other 4 cards can be selected in ways.
Hence, using the multiplication principle, required the number of 5 card combination
...

**Q.5. ** Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

There are 6 red balls, 5 white balls and 5 blue balls.
9 balls have to be selected in such a way that consists of 3 balls of each colour.
3 balls are selected from 6 red balls in .
3 balls are selected from 5 white balls in
3 balls are selected from 5 blue balls in .
Hence, by the multiplication principle, the number of ways of selecting 9 balls
...

**Q.4.** In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

A team of 3 boys and 3 girls be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in ways.
3 girls can be selected from 4 boys in ways.
Therefore, by the multiplication principle, the number of ways in which a team of 3 boys and 3 girls can be selected
...

**Q.3. ** How many chords can be drawn through 21 points on a circle?

To draw chords 2 points are required on the circle.
To know the number of chords on the circle , when points on the circle are 21.
Combinations =Number of chords

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