## Kerala Plus One Maths Chapter Wise Previous Questions and Answers Chapter 7 Permutation and Combinations

Question 1.

a. Find n, if 12 × (n-1) p_{3} = 5 ×(n+1) p_{3 }b If ^{n}p_{r} = 840; ^{n}C_{r} = 35 find r.

c. English alphabet has 5 vowels and 21 consonants. How many 4 letter words with two different vowels and two different consonants can be formed without repetition of letters? [March-2018]

Answer:

a. 12 × (n-1)P_{3} = 5 × (n+l)P_{3 }12 × (n-1)(n-2)(n-3) = 5 × (n+1)n(n-1)

^{5}C_{2} ways .Two different consonents can be selected in ^{21}C_{2} ways.

Total selection 4 letters = ^{5}C_{2} × ^{21}C_{2 }Total words = 4! ×^{5}C_{2} × ^{21}C_{2 }= 10 ×210 × 24 = 50400

Question 2.

a. If then x is …………….

i. 32

ii. 16

iii. 64

iv. 8

b. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags one below the other?

c. Find r if ; ^{5}P =2.^{6}P_{r-1 }**OR**

a. If ^{n}C_{9} = ^{n}C_{8}, then n =………………….

i. 9

ii. 8

iii. 17

iv. 1

b. How many chords can be drawn through 12 points on a circle?

c. What is the number of ways of choosing 4 cards from a pack of 52 playing cards?

In how many of these :

i. Four cards are of the same suit?

ii. Cards are of the same colour? [March-2017]

Answer:

Question 3.

a. Write the value of ^{7}C_{5}.

b. Find the value of n if 3. ^{n}P_{4} = 5.^{(n-1)}P_{4 }c. What is the number of ways of choosing four cards from a pack of 52 cards, provided all four cards belong to four different suits?

**OR **a.

^{29}C

_{29}=…………………

b. Find the value of n, if

12

^{(n-1)}P

_{3}= 5.

^{(n+1)}P

_{3 }c. 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 at least one boy and one girl? [March – 2016]

Answer:

Question 4.

a.

b. Find r,5 × ^{4}P_{r}= 6 × ^{5}P_{r-1}

c. Find the number of 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.

**OR **i. n – 1

ii. n

iii. 0

iv. 1

b. If

^{n}C

_{9}=

^{n}C

_{g}, find

^{n}C

_{2}.

c. How many ways can a team of 5 persons be selected out of a group of 4 men and 7 women, if the team has atleast one man and one woman?

Answer:

c. 8 letter arrangements = 8 ! = 40320

i. All vowels occur together = 6! x 3 !

DAUGHTER

AUEDGHTR; =40320 ways

ii. Vowels are not occur together

=40320-4320 = 36000

Question 5.

a.In how many ways can the letters of the word, PERMUTATIONS be arranged if:

i. The words start with P and end with S?

ii. There are always 4 letters between P and S?

b. In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together.

c. How many chords can be drawn through 21 points? [March-2014]

Answer:

a. i. Total 12 letters (T is repeated twice), number of words formed

= = 1814400

ii. There are 5 vowels and 7 consonants. When vowels are all together, there are 7 consonants and 1 group of 5 vowels. This an be arranged in wats (T occurs twice).

Also 5 vowels can be arranged in 5! ways. Required number of ways

= 5!× = 2419200

iii. If there are exactly 4 letters between P and S, P and S can occupy 1^{st} and 6^{th} places or 2^{nd} and 7^{th} places or 3^{rd} and 8^{th} places, or 4^{th }and 9^{th} places or 5^{th} and 10^{th} places or 6^{th }and 11^{th} places or 7^{th} and 12^{th} places (7 ways).

Also P and S can interchange their positions. Thus P and S can be arranged in 14 ways remaining 10 places can be filled in ways (T occur twice)

Number of words

b. Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can

be seated only at the cross marked places. ×G×G×G×G×G×

There are 6 cross marked places and the three boys can be seated in ^{6}P_{3} ways.Hence, by multiplication principle, the total number of ways.

5! ×^{6}C_{3}= 5! ×

4 ×5 ×2 ×3 ×4 ×5 ×6 = 14400

Question 6.

a. Find the value of n such that 3.^{n}P_{4}=5.^{n–}^{1}P_{4}, n>4

b. In how many ways can 5 students be seated on a bench?

c. Find the number of different 8 – letters arrangements that can be made from the letters of the word, ‘DAUGHTER’ so that:

i. All vowels occur together,

ii. All vowels do not occur together.

**OR**

a. Determine n if ^{2n} C_{3} = 11.^{n}C_{3}.

b. In how many ways can a cricket team of 11 players be selected from 15 players?

c. A bag contains 5 white, 6 red and 4 blue balls. Determine the number of way in which 2 white, 3 red and 2 blue balls can be selected. [March-2013]

Answer:

.

b. 5 students can be seated on a bench in 5 ways, i.e., 5! = 120

c. 8 letter arrangements = 8 ! = 40320

i. All vowels occur together = 6! x 3 !

DAUGHTER

AUEDGHTR; =40320 ways

ii. Vowels are not occur together

=40320-4320 = 36000

c. No. of ways to select 2 whiteballs from 5 white balls = 5C_{2 }No. of ways to select 3 red balls from 6 red balls = 6C_{3 }No. of ways to select 2 blue balls from 4 blue balls = 4 C_{2 }Total No. of ways = 5 C_{2}×6C_{3}×4C_{2}= 1200

Question 7.

i. Find n, if ^{n}P_{5} = 42 ^{n}P_{3}, for n > 4,

ii. In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?

**OR **i. Find n if 3

^{n}P

_{4}= 5

^{(n–}

^{1)}P

_{4}, for n > 4.

ii. A committee of 3 persons is to be constituted from a group of 2 men and 3 women. How many of these committees would consist of a man and 2 women? [February – 2013]

Answer:

∴ n=10 or -3

here n>4,∴n=10.

ii. Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can

be seated only at the cross marked places. ×G×G×G×G×G×

There are 6 cross marked places and the three boys can be seated in

^{6}P

_{3}ways.Hence, by multiplication principle, the total number of ways.

5! ×

^{6}C

_{3}= 5! ×

4 ×5 ×2 ×3 ×4 ×5 ×6 = 14400

ii. Total ways = 2× 3 C

_{2}= 6 ways

Question 8.

a. Solve for the natural number n;12. ^{n(n-1)}P_{3 }= 5. ^{(n + 1)}P_{3 }b. In how many ways can 7 athletes be choose out of 12?

c. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed without repetition of letters?

**OR**

a. Find r if ^{5}P_{r} =^{6}P_{r-1 }b. If there are 12 persons in a party and each of them shake hands with all others, what is the total number of handshakes?

c. In how many ways can a committee of 3 men and 2 women be selected out of 7 men and 5 women? [March-2012]

Answer:

Question 9.

i. Simplify

ii. In how many different ways can the letters of the word HEXAGON be permuted?

iii. In how many different ways can a team of 3 boys and 3 girls be seleted from 5 boys and 4 girls?

**OR**

i. then find x.

ii. How many 4 digit numbers are there with no digit repeated?

iii. If ^{n}C_{8} =^{n}C_{2}, then find nC_{3}? [March-2011]

Answer:

Question 10.

i. Find the value of n such that nP_{5} = 42 nP_{3 }for n > 4

ii. A committee of 3 persons is to be constituted from a group of 2 men and 3 women.

a. In how many ways can this be done?

b. How many of these committees would consist of 1 man and 2 women.

**OR **i. If

^{n}C

_{2}–

^{2n}C

_{1 }find n.

ii.

a. Find the number of words that can be formed from the letters of the world MALAYALAM.

b. How many of these arrangements start with Y? [March-2010]

Answer:

∴ n=10 or -3

here n>4,∴n=10.

ii. Total ways = 2× 3 C

_{2}= 6 ways

Question 11.

a. If ^{4}P_{r} = 6×^{5}P_{r-1} .Find ‘r’.

b. How many 3 digit numbers can be formed with the digits 0,1,2,3, and 4?

c. In a Panchayath there are 10 Panchayath members. Ladies contested only in the 50% reserved constituency. If the post of President and Vice President are reserved for ladies, in how many ways both the President and Vice President can be selected? [September-2010]

OR

a. Prove that ^{n}C_{r} =^{n}C_{r-1 }b. Twenty eight matches were played in a volley ball tournament. Each team playing one against each of others. How many teams were there?

c. If the letters of the word “TUTOR” be permuted among themselves and arranged as in a dictionary, then find the position of the word TUTOR

Answer:

Question 12.

i. Find ‘r’ if ^{5}P_{r} =^{6}P_{r-1 }ii. How many 4 digit numbers are there with distinct digits?

**OR**

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

ii. If 2 ^{n}C_{3}: ^{n}C_{3} = 11:1, then find the value of n. [August-2009]

Answer:

OR

i.

Question 13.

i. Find n if ^{(n-1)}p: ^{n}P = 1 : 9

ii. 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 from the lot. [March-2009, September-2008]

Answer:

Question 14.

i. Find r if ^{5}P_{r} = ^{6}P_{r-1 }ii. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination. [March-2009]

Answer:

ii One ace card can be selected from 4 ace cards in 4 ways. Remaining 4 cards are selected from 48 cards which can be done in 48 C_{4} ways.

Question 15.

Find r if ^{10}P_{r}=2.^{9}P_{r} [September-2008]

Answer:

Question 16.

i. A team of 8 players is to be chosen from a group of 12 players. One of the 8 is then selected as captain and another as vice caption. In how many ways can the team be selected?

ii. A box contains 4 red balls 3 white balls and 5 black balls. In how many different ways can a red and a black ball be selected. [June-2008]

Answer:

i. Number of ways of selecting the team

Number of ways of selecting a captain and a vice captain (2 persons) from 8 players = 8P_{2} = 8× 7 = 56

By fundamental principle of counting, total number of ways = 495 ×56 = 27720

ii 4C_{1}×5C_{1} = 4×5= 20

Question 17.

i. ^{10}C_{4}+ ^{10}C_{5 }ii. How many words can be made from the letters in the word MONDAY without repetition if 4 letters are used at a time. [June-2007, February -2008]

Answer:

i. ^{10}C_{4} +^{10}C_{5} =^{11}C_{5 }nC_{r} + nC_{r-1} = n+1C_{r})

ii There are 6 letters in MONDAY

∴ Number of words = 6 P = 360

Question 18.

The letters of the word TUESDAY are arranged in a line each arrangement ending with letter S.

i. How many different arrangement are possible.

ii. How many of them starts with letter D. [March-2006]

Answer:

i. There are 7 letters in TUESDAY since each arrangement ends with S,

number of arrangements = 6! = 720

ii. Number of arrangements starting with D and ending in S = number of arrangements of 5 letters

= 5! = 120

Question 19.

A mathematics paper consists of 10 questions divided into two parts I and n, each part containing 5 questions. A student is required to attempt 6 question in all, taking at least two question from each part

i. What are the possible choices?

ii. In how many ways can the student select the question. [September-2007]

Answer:

i. The student can select the questions as shown below:

Part I | Part II | No. of ways |

2 | 4 | 5C_{3}×5C_{4} =50 |

3 | 3 | 5 C_{3}×5 C_{3}= 100 |

4 | 2 | 5 C_{4}×5 C_{2}= 100 |

if Total number of ways of selection = 50+100 + 50 = 200

Question 20.

If the letters of the word ‘FATHER’ be permuted and arranged as in a dictionary, find the rank of the word. [June-2007]

Answer:

The letters of the word ‘FATHER’ when arranged in alphabetical order is AEFHRT. The different possible arrangements of the letters are given below:

A——– = 1.(5.4.3.2.1) =120

E———–= 1.(5.4.3.2.1) = 120

FAE——- =1.1.1.(3.2.1) =6

FAH—— = 1.1.1.(3.2.1) =6

FAR——- = l.Ll.(3.2.1) =6

FATE—– = 1.1.1.1.(2.1) = 2

FATHE — =1.1.1.1.1.(1)=1

Rank of FATHER

= 120+120 + 6 + 6 + 6 + 2 +1 =261

Question 21.

i. In how many ways can a committee selected from 15 persons if the committee is to have 3 members.

ii. What is the relation between ^{n}P_{r} and ^{n}C_{r-1} ? Show that ^{n}C_{r}+^{n}C_{r-1} = ^{n}^{+1}C_{r} [June-2007]

Answer:

Question 22.

i. Using the alphabets of the word BANANA, determine how many words can be formed so that the two N’s are not together.

ii. Seven different points on the circumference of a circle are chosen. These points are joined pairwise by straight lines.

How many straight lines can be drawn. [February -2008]

Answer:

i. There are 6 letters (3A’s, 2N’s and B)

Total no. of arrangements =

Consider the 2 N’s as a single unit. Then there will be 5 units.

∴ no. of words with N’s together =

∴ no. of words with N’s together = 60-20=40

ii. Any two points on a circle can be joined by a straight line

No.of straight lines = 7C_{2 }=