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Q1. There are three groups of men in a company having 4, 5 and 6 members respectively. Two men have to be selected from the same group as President and Vice-President of a committee to be formed. Find the number of ways in which such a selection can be made.

Solution

Let G1, G2, G3 denote the three groups. Two men as President and Vice-President can be selected from G1 in 43 = 12 ways. The same selection can be made from G2 in 5  4 = 20 ways and from G3 in 6  5 = 30 ways. Since these selections are mutually disjoint, by fundamental principle of addition, the total number of ways the selection can be made is 12 + 20 + 30 = 62.

Q2. If 8 parallel lines in a plane are intersected by a family of 10 parallel lines. Find the number of parallelograms formed?

Solution

A parallelogram is formed by choosing two straight lines 2 lines from the set of 8 lines and 2 lines from the set of 10 parallel lines. Two parallel lines can be chosen in  ways from the set of 8 parallel lines and 2 parallel lines from the set of 10 parallel lines can be chosen in  ways.  Hence, the number of parallelogram formed =

Q3. How many triangles can be drawn through 10 points in a plane?

Solution

A triangle requires three points to be drawn, hence the total number of triangles is    ¹⁰ C₃  

Q4. Find the number of ways of selecting 9 balls from 6 red balls, 5 black balls and 5 green balls if each selection consists of 3 balls of each color, assuming that the balls of same color are distinguishable.

Solution

Since balls of same color are such that we can distinguish them, therefore we have 6 different red balls, 5 different black balls and 5 different green balls.   Here we have to select 3 balls of each color and so a total of 9 balls.    The number of ways of selecting 3 red balls from 6 different red balls.                                                                             The number of ways of selecting 3 black balls from 5 different black balls                                                                                  Similarly the number of ways of selecting 3 green balls out of 5 different green balls                                                                                                  Hence, the required number of selecting 9 balls = 20 x 10 x 10 = 2000 ways

Q5. How many words (with or without meaning) of three distinct letters of the English alphabets are there?

Solution

Here we have to make 3 letter words.  That means we have to fill up three places by distinct letters of the English alphabets.   Since there are 26 letters of the English alphabet, the first place can be filled by anyone of these letters.  So, there are 26 ways of filling up the first place.  Now the second place can be filled up by anyone of the remaining 25 letters.  So, there are 25 ways of filling up the second place.  After filling up the first two places only 24 letters are left to fill up the third place.  So the third place can be filled in 24 ways.   Hence the required number of words = 26 x 25 x 24 = 15600

Q6. In how many ways 5 rings of different types can be worn in 4 fingers?

Solution

The first ring can be worn in any of the 4 fingers. So there are 4 ways of wearing it. Similarly each one of the other rings can be worn in 4 ways. Hence, the required number of ways = 4 x 4 x 4 x 4 x 4 = 4⁵

Q7. How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

Solution

Here we are available with 7 – digits, one 1, three 2’s, one 0 and two 4’s.  Also ,the number 1000000 has 7 digits and we want to form numbers greater than 1000000.     So, the required numbers should also be of 7 digits. _{The number of all type of arrangements possible with the given digits is                                         Now these arrangements also include the cases when  0 is at extreme left position which makes the number a 6 digit one. The number of such numbers is                                                 Excluding these numbers, we get the required number of numbers = 420 – 60 = 360.}

Q8. Twelve students compete in a race. In how many ways can the first three places be taken?

Solution

No. of students = 12 No. of ways can the first three places is taken is equal to No. of ways of arranging 12 students taking 3 at a time.

Q9. There are 15 points in a plane, no more than 2 of which are collinear. How many pentagons in all can be made using these points?

Solution

A pentagon requires 5 points. Hence, the total number of pentagons will be   Hence 3003 pentagons are possible in all.

Q10. 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 (ii) At least 3 girls (iii) At most 3 girls

Solution

(i) A committee consisting of 3 girls and 4 boys can be formed in ways =504 ways   (ii) A committee having at least 3 girls will consists of   (a) 3 girls 4 boys (b) 4 girls 3 boys   This can be done in  ways   (iii) A committee having at most 3 girls will consist of (a) No girl 7 boys (b) 1 girl, 6 boys (c) 2 girls, 5 boys (d) 3 girls, 4 boys  

Q11. A library membership allows 4 books to be borrowed at any given time. Rahul goes to the library and finds that he likes two books on a given shelf which contains 15 books in all. Rahul has decided that he would either take both the books or take none of these two. In how many ways can he select 4 books?

Solution

Rahul can either select both the books and select 2 out of the other 13 this can be done in ¹³C₂ ways. Rahul can select 4 books leaving these 2 out which can be done in ¹³C₄ ways. Hence total no of selections=

Q12. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

Solution

_{The word ASSASSINATION has three A's, four S's, two I's, two N's and T, O occurs only once. We have to consider the case when all the S's together and so taking it as one packet or unit. So now we have three A's, one unit of four S's, two I's, two N's, one T, one O and thus a total of 10 units. Therefore the number of arrangements possible when all the S's is together                nbsp;             Hence, the distinct permutations of the letters of the word ASSASSINATION when four S's  come together = 151200.}

Q13. Code is to be formed by taking 2 letters from vowels of English alphabets followed by 3 digits chosen from 1, 2, 3 or 4. How many ways this code can be made if repetition is allowed.

Solution

Here vowels can be arranged in 5 5=25 ways _{Digits can be arranged in 444=64 ways So, total arrangements of the codes are 2564=1600.}

Q14. 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 they be chosen? Now a day's teenager's behavior is highly influenced under peer pressure without considering its pros and cons. Do you think that teachers' suggestion plays an important role in development of students' life?

Solution

We have following cases: (i) Three particular students join: now we have to choose 7 students from remaining 22 students So number of ways = ²²C₇ (ii) Three particular students do not join. Here we have to choose 10 students from remaining 22 students So number of ways = ²²C₁₀ Hence, total number of ways =  ^{+ 22C₁₀ = 817190 Teachers' suggestion plays a very important role in development of students' life. They mould the students to bring out their skills or improvise them, teach good habits/attitudes and help them to become good citizens of the nation.}

Q15. How many four digit numbers can be formed using the digits 0,1,2,3,4,5 if 1)     repetition of digits is not allowed? 2)     repetition of digits is allowed?

Solution

1) In the four digit number 0 cannot appear in the thousand’s place. So, thousand’s place can be filled in 5 ways (using 1,2,3,4 or 5).  Since repetition of digits is not allowed and 0 can be used at hundred’s place, so hundred’s place can be filled in 5 ways.  Now, any one of the remaining four digits can be used to fill up ten’s place.  So ten’s place can be filled in 4 ways.  One’s place can be filled from the remaining three digits in 3 ways.   Hence, the required number of numbers = 5 x 5 x 4 x 3 = 300   2)  For a four digit number we have to fill up four places and 0 cannot appear in the thousand’s place can be filled in 5 ways.  Since repetition of digits is allowed, so each of remaining three places hundred’s, ten’s and one’s can be filled in 6 ways.   Hence, the required number of numbers = 5 x 6 x 6 x 6 = 1080.

Q16. In how many can three white balls, four black balls and five red balls be arranged.

Solution

In totality we have 3 + 4 + 5 = 12 balls of which 3 are of one kind, 4 are of second kind and 5 are of third kind. _{Hence, the distinct permutations of the balls}   

Q17. In how many ways 6 children can stand in a queue?

Solution

6 children can stand in a queue in  ways.

Q18. If , then find the value of n.

Solution

We have

Q19. In how many ways can 9 persons sit around a table so that all shall not have the same neighbours in any two arrangements?

Solution

Persons can sit at a round table in  ways. But in clockwise and anti-clockwise arrangement, every person will have the same neighbours. So, the required numbers of ways are

Q20. How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?

Solution

A number between 100 and 1000 contains 3 digits.  So, we have to form 3 digit numbers having exactly one of their digits as7. Such type of numbers can be divided into three types:   1)     There numbers that have 7 in the unit place but not in any other place. 2)     These numbers that have 7 in the ten's place but not in any other place. 3)     These numbers that have 7 in the hundred's place but not in any other ways.   We shall now count these three type of numbers separately.   1)  These 3-digit numbers that have 7 in the unit's place but not in any other place.   The hundred's place can have any one of the digits from 0 to 9 except 0 and 7.  So hundred's place can be filled in 8 ways.  The ten's place can have anyone of  the digits from 0 to 9 except 7.  So the number of ways the ten's place can be filled is 9.  The unit's place has 7.  So, it can be filled in only one way.   Thus, there are 8 x 9 x 1 = 72 numbers of the first kind.   2)  Those three-digit numbers that have 7 in the ten's place but not in any other place.   The number of ways to fill the hundred's place = 8 [By any one of the digits from                                                                                             1,2,3,4,5,6,8,9] The number of ways to fill the ten's place          = 1 [By 7 only] The number of ways to fill the one's place         =  9 [By any one of the digits 0,1,2,3,4,5,                                                                                  6,8,9] Thus, these are 8 x 1 x 9 = 72 numbers of the second kind.   3)  These three digit numbers that have 7 in the hundred's place but not at any other place.   In this case, the hundred's place can be filled only in one way and each of the ten's and one's place can be filled in 9 ways.  So, there are 1 x 9 x 9 = 81 numbers of the third kind. Hence, the total number of required type of numbers = 72 + 72 + 81 = 225.