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

**Short Answer** **Type Questions **

**(Score 3)**

Question 1.

a. Find the value of

b. Find the value of x if

Answer:

a.

Question 2.

a In a multiple choice question, there are four alternatives, of which one or more are correct. Number of ways in which a candidate can attempt these question is

a. 4P_{4}

b.4

c. 2^{4 }d. 15

b. 7!= ………….

Answer:

a. The total number of ways of ticking the answers in any one attempt = 2^{4}-1=15.

b. 5040

7!=7×6×5×4×3×2×1 = 5040

Question 3.

Number of even integers that can be formed by using all digits 1,2,3,4 and 5 (without repetition) is….

Answer:

Since repetition is not allowed unit place can be filled in 2 ways {2,4} tens place can be filled in 4 ways hundred place can be filled in 3 ways, thousand place can be filled in 2 ways ten thousand place can be filled in 1 ways, total number of ways = 1,2.3.4.2 = 48

Question 4.

a. The number of ways in which r letters can be posted in m letter boxes in a town is

a. mP_{r}

b. nC_{r}

c. m^{r}

d. r^{m }b. The number of ways in which a necklace can be formed by using 5 identical red beads and 6 identical white beads is

a.

b.

c.

d. 11

Answer:

a. m^{r}

Assuming in boxes are different.

Total number of ways in which r letter can be posted = m^{r }

b. In case of necklase, clockwise and anti clockwise arrangement are considered alike, so the number of necklace that can be formed is

Question 5.

Given n points on the circle. Find the number of

i. Lines through the two points.

ii. Triangles

Answer:

i. Through any two points on the circle, a line can be drawn. The number of lines drawn through n points on the circle = ^{n}C_{2 }ii. Any three points on the circle are non col linear. So a triangle can be drawn through the 3 points on the circle.

∴ Number of triangles drawn through n points = ^{n}C_{3}

Question 6.

If ^{12}c_{4} + ^{12}c_{s} = n c_{5}, then find n.

Answer:

Question 7.

The number of ways in which one can select three distinct integers between 1 and 30, both inclusive, whose sum is even is

a. 455

b. 140

c.2030

d.400

Answer:

c. 2030

Number are either all even or one even and other two odd required number of ways =^{15}C_{3} + ^{15}C_{1} × ^{16}C_{2 }

= 455 + (15 × 105) =455 + 1575 =2030

**Short Answer Type Questions**

**(Score 4)**

Question 1.

Consider the word ANNAMALAI,

i. How many new words can be formed using the given word.

ii. Among the new words how many of them will begin with A and end with I.

Answer:

i. ANNAMALAI has 4 A’s and 2 N’s

No. of the new words =

ii. If we fix A and I, no. of A’s =j 3 and number of N’s = 2

∴ No.of words

Question 2.

a. In how many different ways can a pack of 52 cards be dealt among 4 players so that each receives 13 cards.

b. How many numbers with no more than three digits can be formed using only the digits 1 through 7 with no digit used more than once in a given number?

Answer:

Question 3.

a. Prove that n(n-1)(n-2)………. (n – r + 1) =

b. Compare

Answer:

Question 4.

Find the values of the foUowing

i. ^{5}P_{3}

ii. P(15,3)

Answer:

Question 5.

If ^{56}P_{r+6} : ^{54}P_{r+3} = 30800: 1, then find the value of x.

Answer:

Question 6.

a. Find the number of ways in which one can select three distinct integers between 1 and 30, both inclusive, whose sum is even.

a. 455

b. 140

c.2030

d.400

b. Find the total number of 9-digit numbers which have all different digits.

Answer:

a. 2030

Number are either all even or one even and other two odd required number of ways

= 455 + (15 x 105) =455 + 1575 =2030

b. The total number of 9-digit numbers, having all digits are different

=

Question 7.

There are 15 points in a plane, no three of which are collinear, except 4 points which lie on a straight line.

i. How many straight lines can be formed by joining them?

ii. How many triangles can be formed?

Answer:

i. A straight line is drawn through two points. ^{15}C_{2} straight lines can be drawn.

Since 4 points lie on a line, ^{4}C_{2} lines cannot be there. Instead there will be one line with four points.

Total number of lines = ^{15}C_{2}– ^{4}C_{2}+ 1

ii. A triangle is drawn by 3 points.

No. of triangles = ^{15}C_{3 }Since, 4 points are collinear, ^{4}C_{3} triangles cannot be there.

∴ No. of triangles = ^{15}C_{3}– ^{4}C_{3 }

Question 8.

a. If ^{n}C_{2}=^{n}C_{g}, then find n.

b. Find n if ^{n}P_{5} = 42 × ^{n}P_{5};n>4.

Answer:

a. n= 10

b. n(n- 1 )(n-2)(n-3)(n-4) = 42n(n- 1 )(n-2)

(n-3)(n-4) = 42

n2 – 7n 0=0

(n – 10) (n +3) = O

n=10, n=-3

As n cannot be negative, so n = 10

**Long Answer Type Questions**

**(Score – 6)**

Question 1.

Out of 6 teachers and 4 students, a committee of 8 is to be chosen. In how many ways can this be done.

i. When there are not less than 4 teachers in the committee.

ii. When there is a majority of teachers.

Answer:

i. When there are not less than 4 teachers in the committee.

Committee may be chosen in the following ways:

Question 2.

A cricket team of 11 is to be selected out of 15 players. In how many ways can it be done, if

i. A particular ‘player A’ is to be included.

ii. A particular ‘player B’ is to be excluded.

Answer:

Total number of players = 15

Number of players to be selected = 11 L

i. Since the ‘Player A’ is included, we have to select 10 players from the remaining 14 players. This can be done in ^{14}C_{10} ways.

ii. Since the ‘Player B’ is excluded, we have to select 11 players from the remaining 14 players. This can be done in ^{14}C_{10} ways.

No. of ways of selection so that the ‘Player B’ is excluded = ^{14}C_{11}

Question 3.

If ^{n}C_{x}=^{n}C_{y}, then either x = y or x + y = n.

a. If ^{n}C_{10} = ^{n}C_{12}, then find the value of ^{23}C_{n}.

b. Prove that r.^{n}C_{r} = n. ^{n-1}C_{r-1 }Answer:

Question 4.

If ^{n+2}C_{8} ,: ^{n-2}P = 57:16, then find the value of n.

Answer:

Question 5.

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. Four cards belong to four different suits?

iii. Are face cards?

Answer:

Here, we have to choose 4 cards out of 52 playing cards at a time. So,

Required number of ways = ^{52}C_{4}.

i. There are four suits diamond, club, spade, heart in a pack of 52 playing cards, and each suit have 13 cards.

So from each suit, 4 cards can be choosed in ^{13}C_{4} ways separately.

Required number of ways

ii. Four cards belong to four different suits i.e., one card is selected from each suit. So, required number of ways

= ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1 }= 13 × 13×13 × 13 = 13^{4 }iii. There are 12 face cards and we have to select 4 cards out of these.

Required number of ways = ^{12}C_{4 }

**NCERT Questions and Answers **

Question 1.

Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?

Answer:

There will be as many signals as there are ways of filling in 2 vacant places in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals = 4×3 = 12.

Question 2.

How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?Answer:

There will be as many ways as there are ways of filling 2 vacant places in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2× 5, i.e., 10.

Question 3.

How many 4 digit odd numbers can be formed with the help of the digits 1,2,3,4,5 if

i. No digit is repeated.

ii. Digits are repeated.

Answer:

i.The digit in unit place can be 1, 3, 5. This can be done in 3 ways. The other three places can be filled in ^{4}P_{3} ways.

∴ Total numbers = 3 × ^{4}P_{3} = 72

ii The digit in unit place can be filled in three ways. The other three places can be filled by using any of the 5 digits given. This can be done in 5 × 5 × 5 ways.

Total numbers = 3×5×5×5 = 375

Question 4.

In how many ways can the letters of the word PERMUTATIONS be arranged if the

i. Word start with P and end with S.

ii. Vowels are all together.

iii. There are always 4 letters between P and S.

Answer:

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

Question 5.

i. If n P_{r} = 720 and n C_{r} = 120 find r.

ii. How many words each of 3 vowels and 2 consonants can be formed from the letters of the word ‘INVOLUTE’.

Answer:

ii. INVOLUTE has 4 vowels and 4 consonants (N, V, L, T).

3 vowels can be selected in ^{4}C_{3} ways and 2 consonants in ^{4}C_{2} ways.

There are 5 letters in each selection and they can be arranged in 5! ways.

Total number of words = ^{4}C_{3} × ^{4}C_{2} ×5! = 4 ×6 ×120 = 2880

Question 6.

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

Answer:

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 7.

i. Determine n if ^{2n}C_{3}: n C_{2} = 12 :1

ii. 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.

Answer:

ii. 3 red balls can be selected from 6 in ^{6}C_{3 }ways. 3 white balls can be selected from 5 white in 5 C_{3 }ways. Similarly 3 blue balls in 5 C_{3 }ways.

∴ Total number of selections

= ^{6}C_{3}×^{5}C_{3}×^{5}C_{3}=20× 10×10 = 2000

Question 8.

In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?

Answer:

The no.of arrangments