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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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