Question 1: From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?

Answer:

A group of 15 cricket players, a team of 11 players is to be chosen in ways

= ^{15} \rm C_{11} = \frac{15!}{11! 4!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365

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Question 2: How many different boat parties of 8 , consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?

Answer:

Number of different boat parties of 8 , consisting of 5 boys and 3 girls, that can be made from 25 boys and 10 girls

= ^{25} \rm C_{5} \times ^{10} \rm C_{3} = \frac{25!}{5! 20!} \times \frac{10!}{3! 7!} = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1 } \times \frac{10 \times 9 \times 8 }{3 \times 2 \times 1} =  6375600

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Question 3: In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?

Answer:

2 courses are compulsory for every student.

Therefore 3 courses are to be chosen from 7 possible courses.

Hence the number of ways a student can choose 5 courses out of 9 courses if 2 courses are compulsory for every student

= ^{7} \rm C_{3} = \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5 }{6} = 35

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Question 4:  In how many ways can a football team of 11 players be selected from 16 players? How many of these will (i) include 2 particular players? (ii) exclude 2 particular players?

Answer:

Number of ways to select 11 players from 16 players

= ^{16} \rm C_{11} = \frac{16!}{11! 5!} = \frac{16 \times 15 \times 14 \times 13 \times 12 }{5 \times 4 \times 3 \times 2 \times 1 } = 4368

i) If two particular players are always included, then the number of ways

= ^{14} \rm C_{9} = \frac{14!}{9! 5!} = \frac{14 \times 13 \times 12 \times 11 \times 10 }{5 \times 4 \times 3 \times 2 \times 1 } = 2002

ii) If two particular players are always excluded, then the number of ways

= ^{14} \rm C_{11} = \frac{14!}{11! 3!} = \frac{14 \times 13 \times 12 }{ 3 \times 2 \times 1 } = 364

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Question 5: There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees:
(i) a particular professor is included.
(iii) a particular student is excluded.
(ii) a particular student is included.

Answer:

Number of ways committee can be formed

= ^{10} \rm C_{2} \times ^{20} \rm C_{3} = \frac{10!}{2! 8!} \times \frac{20!}{3! 17!} = \frac{10 \times 9}{2} \times \frac{20 \times 19 \times 18}{6} = 51300  

i) Number of ways committee can be formed if a particular professor is included

= ^{9} \rm C_{1} \times ^{20} \rm C_{3} = \frac{9!}{1! 8!} \times \frac{20!}{3! 17!} = \frac{9}{1} \times \frac{20 \times 19 \times 18}{6} = 10260  

ii) Number of ways committee can be formed if a particular student is excluded

= ^{10} \rm C_{2} \times ^{19} \rm C_{3} = \frac{10!}{2! 8!} \times \frac{19!}{3! 16!} = \frac{10 \times 9}{2} \times \frac{19 \times 18 \times 17}{6} = 43605 

iii) Number of ways committee can be formed if a particular student is included

= ^{10} \rm C_{2} \times ^{19} \rm C_{2} = \frac{10!}{2! 8!} \times \frac{19!}{2! 17!} = \frac{10 \times 9}{2} \times \frac{19 \times 18 }{2} = 7695 

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Question 6: How many different products can be obtained by multiplying two or more of the number 3, 5, 7, 11 (without repetition)?

Answer:

The number of ways products can be obtained by multiplying two or more of the number 3, 5, 7, 11

=  ^{4} \rm C_{2} + ^{4} \rm C_{3} + ^{4} \rm C_{4} = 6 + 4 + 1 = 11

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Question 7: From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition; at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?

Answer:

Given 2 girls who won the prizes last year should be included.

Therefore we have to select from 12 boys and 8 girls. Therefore the combinations would be

= ^{12} \rm C_{6} \times ^{8} \rm C_{2} + ^{12} \rm C_{5} \times ^{8} \rm C_{3} + ^{12} \rm C_{4} \times ^{8} \rm C_{4}

= \frac{12!}{6! 6!}   \times \frac{8!}{2! 6!} + \frac{12!}{5! 7!}   \times \frac{8!}{3! 5!}   + \frac{12!}{4! 8!}   \times \frac{8!}{4! 4!}

= 924 \times 28 + 792 \times 56 + 495 \times 70

= 104874 

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Question 8: How many different selections of 4 books can be made from 10 different books, if (i) there is no restriction; (ii) two particular books are always selected (iii) two particular books are never selected?

Answer:

i) Number of ways in which 4 books can be made from 10  different books if there is no restriction = ^{10} \rm C_{4} = \frac{10!}{4! 6!} = \frac{ 10 \times 9 \times 8 \times 7 }{ 4 \times 3 \times 2 \times 1 } = 210 

ii) Number of ways in which 4 books can be made from 10   different books if two particular books are always selected = ^{8} \rm C_{2} = \frac{8!}{2! 6!} = \frac{ 8 \times 7  }{  2 \times 1 } = 28 

iii) Number of ways in which 4 books can be made from 10   different books  if two particular books are never selected = ^{8} \rm C_{4} = \frac{8!}{4! 4!} = \frac{ 8 \times 7 \times 6 \times 5 }{ 4 \times 3 \times 2 \times 1 } = 70 

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Question 9: From 4 officers and 8 soldiers in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer?

Answer:

Number of ways in which from 4 officers and 8 soldiers in how many ways can 6 be chosen if there is no restriction

= ^{12} \rm C_{6} = \frac{12!}{6! 6!} = \frac{ 12 \times 11 \times 9 \times 8 \times 7}{ 6 \times 5 \times 4 \times 3 \times 2 \times 1 } = 924 

i) Number of ways in which from 4 officers and 8 soldiers in how many ways can 6 be chosen to include exactly one officer

= ^{4} \rm C_{1} \times ^{8} \rm C_{5} = \frac{4!}{1! 3!} \times \frac{8!}{5! 3!} = 4 \times \frac{8 \times 7 \times 6 }{3 \times 2 \times 1} = 224 

ii) Number of ways in which from 4 officers and 8 soldiers in how many ways can 6 be chosen to include at least one officer

= ^{4} \rm C_{1} \times ^{8} \rm C_{5} + ^{4} \rm C_{2} \times ^{8} \rm C_{4}+^{4} \rm C_{3} \times ^{8} \rm C_{3}+^{4} \rm C_{4} \times ^{8} \rm C_{2}

= \frac{4!}{1! 3!} \times \frac{8!}{5! 3!} + \frac{4!}{2! 2!} \times \frac{8!}{4! 4!} + \frac{4!}{3! 1!} \times \frac{8!}{3! 5!} + \frac{4!}{4! 0!} \times \frac{8!}{2! 6!}

= \Big( \frac{ 4 \times 8 \times 7 \times 6 }{ 3 \times 2 } \Big) + \Big( \frac{ 4 \times 3 \times 8 \times 7 \times 6 \times 5}{ 2 \times 4 \times 3 \times 2 } \Big) + \Big( \frac{ 4 \times 8 \times 7 \times 6 }{ 3 \times 2 } \Big) + \Big( \frac{ 8 \times 7  }{ 2 \times 1 } \Big)

= (4 \times 8 \times 7 ) + (4 \times 3 \times 7 \times 5 ) + (4 \times 8 \times 7 ) + (4 \times 7 )

= 224 + 420 + 224 + 28 = 896

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Question 10: A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted?

Answer:

The number of ways a sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII from 20 students in each of these classes

= ^{20} \rm C_{5} \times ^{20} \rm C_{6}  + ^{20} \rm C_{6} \times ^{20} \rm C_{5}

= \frac{20!}{5! 15!} \times \frac{20!}{6! 14!} + \frac{20!}{6! 14!} \times \frac{20!}{5! 15!}

= 2 \times \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1 } \times \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1 }

= 2 \times 19 \times 3 \times 17 \times 16 \times 19 \times 17 \times 8 \times 15

= 1201870080

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Question 11: A student has to answer 10 questions, choosing at least 4 from each of \text{Part A} and \text{Part B} . If there are 6 questions in \text{Part A} and 7 in \text{Part B} , in how many ways can the student choose 10 questions?

Answer:

Total number of questions = 10

Questions in \text{Part A} = 6 , Questions in \text{Part B} = 7

The number of ways the student choose 10 questions

= ^{6} \rm C_{4} \times ^{7} \rm C_{6}  + ^{6} \rm C_{5} \times ^{7} \rm C_{5} + ^{6} \rm C_{6} \times ^{7} \rm C_{4}

=  \frac{6!}{4! 2!} \times \frac{7!}{6! 1!} + \frac{6!}{5! 1!} \times \frac{7!}{5! 2!} + \frac{6!}{6! 0!} \times \frac{7!}{4! 3!}

=  \frac{6 \times 5 \times 7 }{2} + \frac{6 \times 7 \times 6 }{2} + \frac{7 \times 6 \times 5 }{3 \times 2}

= 105 + 126 + 35

= 266

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Question 12: In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

Answer:

Total number of questions = 5

Number of questions to be answered = 4

Questions 1 and 2 are however compulsory. Therefore the student has to choose 2 questions from the remainder 3 questions.

Number of ways to choose 2 questions from the remainder 3 questions = ^{3} \rm C_{2} = \frac{3!}{2! 1!} = 3

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Question 13: A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions  from either group. In how many ways can he choose the 7 questions?

Answer:

Total number of questions = 5

Number of questions to be answered = 4

Number of ways can he choose the = 7 questions

= ^{6} \rm C_{2} \times ^{6} \rm C_{5}  + ^{6} \rm C_{3} \times ^{6} \rm C_{4} + ^{6} \rm C_{4} \times ^{6} \rm C_{3} + ^{6} \rm C_{5} \times ^{6} \rm C_{2}

=  \frac{6!}{2! 4!} \times \frac{6!}{5! 1!} + \frac{6!}{3! 3!} \times \frac{6!}{4! 2!} + \frac{6!}{4! 2!} \times \frac{6!}{3! 3!} + \frac{6!}{5! 1!} \times \frac{6!}{2! 4!}

=  \frac{6 \times 5  }{2} \times \frac{6 }{1} +  \frac{6 \times 5 \times 4 }{3 \times 2} \times \frac{6 \times 5  }{ 2} +  \frac{6 \times 5 }{ 2} \times \frac{6 \times 5 \times 4  }{ 3 \times 2} +  \frac{6 }{ 1} \times \frac{6 \times 5  }{ 2}

= 105 + 126 + 35

= 266

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Question 14: There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.

Answer:

Number of straight lines that can be drawn within 10 points, taking 2 points at a time = ^{10} \rm C_{2} = \frac{10!}{2! 8!} = \frac{10 \times 9}{ 2} = 45

Number of straight lines that can be drawn within 4 collinear points, taking 2 points at a time = ^{4} \rm C_{2} = \frac{4!}{2! 2!} = \frac{4 \times 3}{ 2} = 6

However, the 4 collinear points will form a straight line.

Hence the total number of lines = 45 - 6 + 1 = 40

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Question 15: Find the number of diagonals of (i) a hexagon (ii) a polygon of 16 sides.

Answer:

Note: A polygon has n vertices. When you join two sides, you will either get a diagonal or a side.

i) Number of diagonals of a hexagon with 6 sides

= ^{6} \rm C_{2} - 6 = \frac{6!}{2! 4!} - 6 = \frac{6 \times 5}{ 2} - 6 = 15 - 6 = 9

ii)  Number of diagonals of a polygon with 16 sides

= ^{16} \rm C_{2} - 16 = \frac{16!}{2! 14!} - 16 = \frac{16 \times 15}{ 2} - 16 = 120 - 16 = 104

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Question 16: How many triangles can be obtained by joining 12 points, five of which are collinear ?

Answer:

The number of triangles that can be obtained by joining 12 points, five of which are collinear

= ^{12} \rm C_{3} - ^{5} \rm C_{3} = \frac{12!}{3! 9!} - \frac{5!}{3! 2!} = \frac{12 \times 11 \times 10 }{ 3 \times 2} - \frac{5 \times 4}{ 2} = 220 - 10 = 210

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Question 17: In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected ?

Answer:

The number of ways a a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected

= ^{4} \rm C_{1} \times ^{6} \rm C_{4} + ^{4} \rm C_{2} \times ^{6} \rm C_{3} + ^{4} \rm C_{3} \times ^{6} \rm C_{2} + ^{4} \rm C_{4} \times ^{6} \rm C_{1}

= \frac{4!}{1! 3!}   \times \frac{6!}{4! 2!} + \frac{4!}{2! 2!} \times \frac{6!}{3! 3!} + \frac{4!}{3! 1!} \times \frac{6!}{2! 4!} + \frac{4!}{4! 0!} \times \frac{6!}{1! 5!}

= \frac{4 }{1} \times \frac{6 \times 5}{ 2 \times 1} + \frac{4 \times 3 }{ 2 \times 1} \times \frac{6 \times 5 \times 4  }{  3 \times 2 \times 1} + \frac{4 }{1} \times \frac{6 \times 5}{ 2 \times 1} +  6

= 60+120+60+6 = 246

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Question 18: In a village, there are 87 families of which 52 families have at most 2 children. In a rural development program, 20 families are to be helped chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made ?

Answer:

There are 52 families that have at most 2 children.

There are 35 families that have more than 2 children.

Selection of 20 families of which at least 18 families must have at most 2 children

= ^{52} \rm C_{18} \times ^{35} \rm C_{2} + ^{52} \rm C_{19} \times ^{35} \rm C_{1} + ^{52} \rm C_{20} \times ^{35} \rm C_{0}

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Question 19: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls?

Answer:

Number of girls = 4

Number of Boys = 7

i) Number of ways 5 members be selected if the team has no girl

= ^{7} \rm C_{5} = \frac{7!}{5! 2!} = \frac{7 \times 6 }{ 2 \times 1} = 21

ii) Number of ways 5 members be selected if the team has at least 1 boy and 1 girls

= ^{4} \rm C_{1} \times ^{7} \rm C_{4} + ^{4} \rm C_{2} \times ^{7} \rm C_{3} + ^{4} \rm C_{3} \times ^{7} \rm C_{2} + ^{4} \rm C_{4} \times ^{7} \rm C_{1}

= \frac{4!}{1! 3!} \times \frac{7!}{4! 3!} + \frac{4!}{2! 2!} \times \frac{7!}{3! 4!} + \frac{4!}{3! 1!} \times \frac{7!}{2! 5!} + \frac{4!}{4! 0!} \times \frac{7!}{1! 6!}

= 140 + 210 + 84 + 7

= 441

iii) Number of ways 5 members be selected if the team has at least 3 girls

= ^{4} \rm C_{3} \times ^{7} \rm C_{2} + ^{4} \rm C_{4} \times ^{7} \rm C_{1}

= \frac{4!}{3! 1!} \times \frac{7!}{2! 5!} + \frac{4!}{4! 0!} \times \frac{7!}{1! 6!}

= 84 + 7

= 91

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Question 20: A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Answer:

i) Number of ways committee can be selected

= ^{5} \rm C_{3} = \frac{5!}{3! 2!} = \frac{5 \times 4 }{ 2 \times 1} = 10

ii) Number of ways committee can be selected if consists of 1 man and 2 women

= ^{2} \rm C_{1} \times ^{3} \rm C_{2} = \frac{2!}{1! 1!} \times \frac{3!}{2! 1!} = 6

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Question 21: Find the number of (i) diagonals (ii) triangles formed in a decagon.

Answer:

Number of sides in a decagon = 10

i) Number of diagonals = ^{10} \rm C_{2} - 10 = \frac{10!}{2! 8!} - 10 = \frac{10 \times 9 }{ 2 \times 1} - 10  = 45 - 10 = 35

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Question 22: Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king ?

Answer:

Total number of cards in a deck = 52

Number of kings in a deck = 4

Therefore number of ways 5 cards combinations out of a deck of 52 cards can be drawn if at least one of the 5 cards has to be a king

= ^{48} \rm C_{1} \times ^{4} \rm C_{4} + ^{48} \rm C_{2} \times ^{4} \rm C_{3} + ^{48} \rm C_{3} \times ^{4} \rm C_{2} + ^{48} \rm C_{4} \times ^{4} \rm C_{1} 

= \frac{48!}{1! 47!} \times \frac{4!}{4! 0!} + \frac{48!}{2! 46!} \times \frac{4!}{3! 1!} + \frac{48!}{3! 45!} \times \frac{4!}{2! 2!} + \frac{48!}{4! 44!} \times \frac{4!}{1! 3!}

= 48 \times 1 + \frac{48 \times 47 }{ 2 \times 1} \times \frac{4  }{ 2 \times 1} + \frac{48 \times 47 \times 46 }{ 3 \times 2 \times 1} \times \frac{4 \times 3  }{ 2 \times 1} + \frac{48 \times 47 \times 46 \times 45 }{ 4 \times 3 \times 2 \times 1} \times \frac{4   }{  1}

= 48 + 4512 + 103776 + 778320

= 886656

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Question 23: We wish to select 6 persons from 8, but if the \text{Person A} is chosen, then \text{Person B} must be chosen. In how many ways can the selection be made ?

Answer:

We need to select 6 persons from 8

Number of ways persons can be selected when \text{Person A} is selected

= ^{6} \rm C_{4} = \frac{6!}{4! 2!} = 15

Number of ways persons can be selected when \text{Person A} is not selected

= ^{7} \rm C_{6} = \frac{7!}{6! 1!} = 7

Hence the total number of ways = 15 + 7 = 22

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Question 24: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer:

Number of ways in which 3 boys and 3 girls be selected from 5 boys and 4 girls

= ^{5} \rm C_{3} \times ^{4} \rm C_{3}   = \frac{5!}{3! 2!} \times \frac{4!}{3! 1!}   = 10 \times 4 = 40

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Question 25: 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 color.

Answer:

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 color

= ^{6} \rm C_{3} \times ^{5} \rm C_{3} \times ^{5} \rm C_{3} = \frac{6!}{3! 3!} \times \frac{5!}{3! 2!} \times \frac{5!}{3! 2!} = \frac{6 \times 5 \times 4  }{ 3 \times 2 \times 1} \times \frac{5 \times 4  }{ 2 \times 1} \times \frac{5 \times 4  }{ 2 \times 1}

= 20 \times 10 \times 10 = 2000 

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Question 26: Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination

Answer:

Total number of cards in a deck = 52

Number of Aces in a deck = 4

Therefore the number of ways in which 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination

= ^{4} \rm C_{1} \times ^{48} \rm C_{4} = \frac{4!}{1! 3!} \times \frac{48!}{4! 44!} = 4 \times \frac{48 \times 47 \times 46 \times 45  }{ 4\times 3 \times 2 \times 1} =  778320

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Question 27: In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer:

Number of ways one can select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers

= ^{5} \rm C_{4} \times ^{12} \rm C_{7} = 5 \times \frac{12!}{7! 5!} = 5 \times \frac{12 \times 11 \times 10 \times 9 \times 8 }{5 \times 4 \times 3 \times 2 \times 1 } = 5 \times 792 = 3960

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Question 28: 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.

Answer:

The number of ways 2 black and 3 red balls can be selected from a bag containing 5 black and 6 red balls

= ^{5} \rm C_{2} \times ^{6} \rm C_{3} = \frac{5!}{2! 3!} \times \frac{6!}{3! 3!} = \frac{5 \times 4}{2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 200

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Question 29: In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer:

The number of ways can a student can choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student

= ^{7} \rm C_{3} = \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

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Question 30: 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 consist of :
(i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?

Answer:

i) Number of ways to select the committee with exactly 3 girls

= ^{9} \rm C_{4} \times ^{4} \rm C_{3} = \frac{9!}{4! 5!} \times \frac{4!}{3! 1!} = \frac{9 \times 8 \times 7 \times 6 }{4 \times 3 \times 2 \times 1 } \times 4 = 504

ii) Number of ways to select the committee with at least 3 girls

= ^{9} \rm C_{4} \times ^{4} \rm C_{3} + ^{9} \rm C_{3} \times ^{4} \rm C_{4}

= \frac{9!}{4! 5!} \times \frac{4!}{3! 1!} + \frac{9!}{3! 6!} \times \frac{4!}{4! 0!}

= \frac{9 \times 8 \times 7 \times 6 }{4 \times 3 \times 2 \times 1 } \times 4 + \frac{9 \times 8 \times 7 }{ 3 \times 2 \times 1 } \times 1

= 504+84

= 588

ii) Number of ways to select the committee with at most 3 girls

= ^{9} \rm C_{7} \times ^{4} \rm C_{0} + ^{9} \rm C_{6} \times ^{4} \rm C_{1} + ^{9} \rm C_{5} \times ^{4} \rm C_{2} + ^{9} \rm C_{4} \times ^{4} \rm C_{3}

= \frac{9!}{7! 2!} \times \frac{4!}{0! 4!} + \frac{9!}{6! 3!} \times \frac{4!}{1! 3!} + \frac{9!}{5! 4!} \times \frac{4!}{2! 2!} + \frac{9!}{4! 5!} \times \frac{4!}{3! 1!}

= \frac{9 \times 8}{2 \times 1} \times 1 + \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 4 + \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} + \frac{9 \times 8 \times 7 \times 6 \times 5}{4\times 3\times 2\times 1 } \times 4

= 36 + 336+756+504

= 1632

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Question 31: 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?

Answer:

Number of ways the student can select the questions

= ^{5} \rm C_{3} \times ^{7} \rm C_{5} + ^{5} \rm C_{4} \times ^{7} \rm C_{4} + ^{5} \rm C_{5} \times ^{7} \rm C_{3}

= \frac{5!}{3! 2!} \times \frac{7!}{5! 2!} + \frac{5!}{4! 1!} \times \frac{7!}{4! 3!} + \frac{5!}{5! 0!} \times \frac{7!}{3! 5!}

= \frac{5 \times 4}{2\times 1} \times \frac{7 \times 6}{2 \times 1} + 5 \times \frac{7 \times 6 \times 5 }{3 \times 2 \times 1 } + 1 \times \frac{7 \times 6}{3 \times 2 \times 1}

= 210 + 175 + 35

= 420

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Question 32: A parallelogram is cut by two sets of m lines parallel to its sides. Find the number of parallelograms thus formed.

Answer:

In a parallelogram, there are two sets of parallel lines.

Now, each set of parallel lines are equal to (m+2) lines.

Hence the number of parallelograms = ^{m+2} \rm C_{2} \times ^{m+2} \rm C_{2} = (^{m+2} \rm C_{2} )^2

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Question 33: Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many (i) straight lines (ii) triangles can be formed by joining them ?

Answer:

Number of triangles formed

= ^{18} \rm C_{3} - ^{5} \rm C_{3} = \frac{18!}{3! 15!} - \frac{5!}{3! 2!} = \frac{18 \times 17 \times 16 }{3 \times 2 \times 1 } - \frac{5 \times 4}{2 \times 1 } = 816 - 10 = 806