Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type

Question 1.

If \({ }^{\mathrm{n}} \mathrm{P}_4\) = 1680, find n. [Mar.’14, May ’06]

Solution:

Given \({ }^{\mathrm{n}} \mathrm{P}_4\) = 1680

\({ }^{\mathrm{n}} \mathrm{P}_4\) = 8 . 7 . 6 . 5

= \({ }^8 \mathrm{P}_4\)

⇒ n = 8

Question 2.

If \({ }^{12} \mathrm{P}_{\mathrm{r}}\) = 1320, find r. [March ’09, AP – May 2015]

Solution:

Given \({ }^{12} \mathrm{P}_{\mathrm{r}}\) = 1320

\({ }^{12} \mathrm{P}_{\mathrm{r}}\) = 12 . 11 . 10

= \({ }^{12} \mathrm{P}_3\)

r = 3.

Question 3.

If \({ }^{(n+1)} P_5:{ }^n P_5\) = 3 : 2, find n.

Solution:

Given \({ }^{(n+1)} P_5:{ }^n P_5\) = 3 : 2

\(\frac{(n+1) P_5}{{ }^n P_5}=\frac{3}{2}\)

⇒ 2n + 2 = 3n – 12

⇒ n = 14.

Question 4.

If \({ }^n \mathbf{P}_7\) = 42 . \({ }^n \mathbf{P}_5\), find n. [TS – Mar. 2017, ’15; May ’12, ’11, ’09, ’07]

Solution:

Given \({ }^n \mathbf{P}_7\) = 42 . \({ }^n \mathbf{P}_5\)

n (n – 1) (n – 2) (n – 3) (n – 4) (n – 5) (n – 6) = 42 . n(n – 1) (n – 2) (n – 3) (n – 4)

(n – 5) (n – 6) = 42

⇒ n^{2} – 11n + 30 – 42 = 0

⇒ n^{2} – 11n – 12 = 0

⇒ n – 12n + n – 12 = 0

⇒ n (n – 12) + 1 (n – 12) = 0

⇒ (n – 12) (n + 1) = 0

⇒ n = 12; n = – 1

Since, n is a positive integer, n = 12.

Question 5.

If \({ }^{56} \mathbf{P}_{(r+6)}:^{54} P_{(r+3)}\) = 30800: 1, find ‘r’.

Solution:

Given \({ }^{56} \mathbf{P}_{(r+6)}:^{54} P_{(r+3)}\) = 30800 : 1

\(\frac{\frac{56 !}{(56-r-6) !}}{\frac{54 !}{(54-r-3) !}}=\frac{30800}{1}\)

⇒ \(\frac{\frac{56 !}{(50-r) !}}{\frac{54 !}{(51-r) !}}=\frac{30800}{1}\)

⇒ \(\frac{\frac{56 \cdot 55 \cdot 54 !}{(50-r) !}}{\frac{54 !}{(51-r)(50-r) !}}=\frac{30800}{1}\)

⇒ 56 55(51 – r) = 30800

⇒ 51 – r = 10

⇒ r = 41.

Question 6.

If \({ }^{12} \mathbf{P}_5+5 \cdot{ }^{12} \mathbf{P}_4={ }^{13} \mathbf{P}_{\mathrm{r}}\), find ‘r’. [TS – May 2015]

Solution:

Given \({ }^{12} \mathbf{P}_5+5 \cdot{ }^{12} \mathbf{P}_4={ }^{13} \mathbf{P}_{\mathrm{r}}\)

\({ }^{(13-1)} P_5+5{ }^{(13-1)} P_{(5-1)}={ }^{13} P_r\)

We know that,

\({ }^n P_r={ }^{(n-1)} P_r+r^{(n-1)} P_{r-1}\)

∴ r = 5

(or) Given \({ }^{12} \mathrm{P}_5+5 \cdot{ }^{12} \mathrm{P}_4={ }^{13} \mathrm{P}_{\mathrm{r}}\)

12 . 11 . 10 . 9 . 8 + 5 . 12 . 11 . 10 . 9 = \({ }^13 \mathrm{P}_{\mathrm{r}}\)

95040 + 59400 = \({ }^13 \mathrm{P}_{\mathrm{r}}\)

⇒ 154440 = \({ }^13 \mathrm{P}_{\mathrm{r}}\)

13 . 12 . 11 . 10 . 9 = \({ }^13 \mathrm{P}_{\mathrm{r}}\)

⇒ \({ }^{13} \mathrm{P}_5\) = \({ }^13 \mathrm{P}_{\mathrm{r}}\)

⇒ r = 5.

Question 7.

If \({ }^{\mathrm{n}} \mathrm{C}_4\) = 210, find n. [TS – Mar.2019]

Solution:

Given \({ }^{\mathrm{n}} \mathrm{C}_4\) = 210

⇒ \(\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}\) = 210

⇒ n (n – 1) (n -2) (n -3) = 5040

n (n – 1) (n – 2) (n – 3) (n – 4) = 10 . 9 . 8 . 7

∴ n = 10.

Question 8.

If \({ }^{12} \mathrm{C}_{\mathrm{r}}\) = 495, find the possible values of ‘r’.

Solution:

Given \({ }^{12} \mathrm{C}_{\mathrm{r}}\) = 495 = 11 . 9 . 5

= \(\frac{11 \cdot 10 \cdot 9 \cdot 5}{10}=\frac{11 \cdot 10 \cdot 9}{1 \cdot 2}=\frac{12 \cdot 11 \cdot 10 \cdot 9}{1 \cdot 2 \cdot 12}\)

= \(\frac{12 \cdot 11 \cdot 10 \cdot 9}{1 \cdot 2 \cdot 3 \cdot 4}={ }^{12} \mathrm{C}_4 \text { (or) }{ }^{12} \mathrm{C}_8\)

∴ r = 4 (or) 8.

Question 9.

If 10. \({ }^n c_2\) = 3 . \({ }^{n+1} C_3\), find ’n’. [May ’12], [AP – Mar. 2015]

Solution:

Given 10. \({ }^n c_2\) = 3 . \({ }^{n+1} C_3\)

⇒ 10 . \(\frac{\mathrm{n}(\mathrm{n}-1)}{1 \cdot 2}\) = 3. \(\frac{(n+1) n(n-1)}{1 \cdot 2 \cdot 3}\)

⇒ n + 1 = 10

⇒ n = 9.

Question 10.

If \({ }^{{ }^n} P_{\mathbf{r}}\) = 5040 and \({ }^n C_{\mathbf{r}}\) = 210 find n and r. [AP – Mar. ‘17, ‘16; Board Paper]

Solution:

\({ }^{{ }^n} P_{\mathbf{r}}\) = 5040 ……………(1)

\({ }^n C_{\mathbf{r}}\) = 210 ……………..(2)

\(\frac{(2)}{(1)} \Rightarrow \frac{{ }^n C_r}{{ }^n P_r}=\frac{210}{5040}\)

\(\frac{\frac{n !}{(n-r) ! r !}}{\frac{n !}{(n-r) !}}=\frac{1}{24} \Rightarrow \frac{1}{r !}=\frac{1}{4 !}\)

r = 4

Now, substituting r = 4 in equation (1)

\({ }^n \mathrm{p}_4\) = 5040

\({ }^n \mathrm{p}_4\) = 10 . 9 . 8 . 7

\({ }^n \mathrm{p}_4={ }^{10} \mathrm{P}_4\)

n = 10.

Question 11.

If \({ }^n C_4={ }^n C_6\), find ‘n’.

Solution:

Given \({ }^n C_4={ }^n C_6\)

If \({ }^n C_r={ }^n C_s\)

⇒ n = r + s (or) r = s

Now, n = r + s

⇒ n = 4 + 6

⇒ n = 10.

Question 12.

If \({ }^{15} \mathrm{C}_{2 \mathrm{r}-1}={ }^{15} \mathrm{C}_{2 \mathrm{r}+4}\) find ’r’. [Mar. ’14, ’05]

Solution:

Given \({ }^{15} \mathrm{C}_{2 \mathrm{r}-1}={ }^{15} \mathrm{C}_{2 \mathrm{r}+4}\)

If \({ }^n C_r={ }^n C_s\)

⇒ n = r + s (or) r = s

⇒ 2r – 1 = 2r + 4

⇒ – 1 ≠ 4

It is not possible.

n = r + s

⇒ 15 = 2r – 1 + 2r + 4

⇒ 15 = 4r + 3

⇒ 4r = 12

⇒ r = 3.

Question 13.

If \({ }^{12} \mathrm{C}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}-5}\), find ‘r’. [TS- Mar. 2016; March ’08]

Solution:

Given \({ }^{12} \mathrm{C}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}-5}\)

If \({ }^n C_r={ }^n C_s\)

⇒ n = r + s (or) r = s

⇒ r + 1 = 3r – 5

⇒ 2r = 6

⇒ r = 3

(or)

n = r + s

⇒ 12 = r – 1 + 3r – 5

⇒ 12 = 4r – 4

⇒ 4r = 16

⇒ r = 4.

∴ r = 3 (or) 4.

Question 14.

If \({ }^9 C_3+{ }^9 C_5={ }^{10} C_{r^{\prime}}\) then find ‘r’.

Solution:

Given \({ }^9 C_3+{ }^9 C_5={ }^{10} C_{r^{\prime}}\)

Question 15.

Find the number of ways of forming a committee of 5 members from 6 men and 3 ladies.

Solution:

Number of ways of forming a committee of 5 members from 6 men and 3 ladies is

\({ }^9 \mathrm{C}_5=\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}\) = 126.

Question 16.

If \({ }^n C_5={ }^n C_6\) then find \({ }^{13} \mathrm{C}_{\mathrm{n}}\). [AP – Mar.2019] [TS-Mar. ‘18; May’14, ‘10, March ’13].

Solution:

Given \({ }^n C_5={ }^n C_6\)

If \({ }^{13} \mathrm{C}_{\mathrm{n}}\)

r = s (or) n = r + s

n = r + s = 5 + 6 =11

Now,

\({ }^{13} C_n={ }^{13} C_2\)

= \({ }^{13} \mathrm{C}_2=\frac{13 \cdot 12}{2 \cdot 1}\)

= 13 . 6 = 78.

Question 17.

Prove that \({ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_6={ }^{11} \mathrm{C}_4\).

Solution:

Given \({ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_6={ }^{11} \mathrm{C}_4\)

L.H.S: \({ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_6={ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4\)

[∵ \({ }^n C_r={ }^n C_{n-r}\)]

= \(={ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_{4-1}={ }^{(10+1)} \mathrm{C}_4\)

[∵ \({ }^n C_r+{ }^n C_{n-r}={ }^{(n+1)} C_r\)]

= \({ }^{11} \mathrm{C}_4\)

= R.H.S

Question 18.

If \({ }^{12} C_{s+1}={ }^{12} C_{2 s-5}\) find ‘s’. [Mar. ’11]

Solution:

Given \({ }^{12} C_{s+1}={ }^{12} C_{2 s-5}\)

If \({ }^n C_r={ }^n C_s\)

⇒ n = r + s (or) r = s

⇒ r = s

⇒ s + 1 = 2s – 5

⇒ s = 6 (or)

n = r + s

12 = s + 1 + 2s – 5

12 = 3s – 4 = 3s = 16

s = \(\frac{16}{3}\)

Since s is integer, s = 6.

Question 19.

If \({ }^n C_{21}={ }^n C_{27}\) find \({ }^{50} \mathrm{C}_{\mathrm{n}}\).

Solution:

Given \({ }^n C_{21}={ }^n C_{27}\)

If \({ }^n C_r={ }^n C_s\)

⇒ n = r + s (or) r = s

⇒ n = r + s

⇒ n = 21 + 27

⇒ n = 48

Now, \({ }^{50} \mathrm{C}_{\mathrm{n}}={ }^{50} \mathrm{C}_{48}\)

= \({ }^{50} \mathrm{C}_2=\frac{50 \cdot 49}{2 \cdot 1}\)

= 25 . 49 = 1225.

Question 20.

Find the number of positive divisors of 1080. [AP – May, Mar. 2016; May ‘13]

Solution:

1080 = 2^{3} × 3^{3} × 5^{1}

The number of positive divisors of 1080 = (3 + 1) (3 + 1) (1 + 1)

= 4 . 4 . 2 = 32.

Question 21.

Find the value of \({ }^{10} \mathrm{C}_5+2 \cdot{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_3\). [AP – Mar. ‘18; TS- Mar. 2017; March ‘10]

Solution:

Given \({ }^{10} \mathrm{C}_5+2 \cdot{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_3\)

= \(\left({ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_4\right)+\left({ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_3\right)\)

= \({ }^{11} C_5+{ }^{11} C_4\)

= \({ }^{12} \mathrm{C}_5=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}\)

= 792

Question 22.

Find the number of injections ofaset A with 5 elements to a set B with 7 elements.

Solution:

If a set A has m elements and the set B has n elements, then the number of injections from A into B is \(\mathrm{n}_{\mathrm{m}}\) if m ≤ n and 0 if m > n.

Given n = 7, m = 5

∴ The number of injections from A to B is \({ }^7 \mathrm{P}_5\) = 7. 6 . 5 . 4 . 3 = 2520.

Question 23.

Find the number of ways in which 4 letters can be put in 4 addressed envelopes so that no letter goes into the envelope meant for it. [TS -May 2016]

Solution:

The number of derangements of n distinct things is

\(\mathrm{n} !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \ldots+(-1)^{\mathrm{n}} \frac{1}{\mathrm{n} !}\right)\)

Required number of ways is \(4 !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)\) = 12 – 4 + 1 = 9.

Question 24.

A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons in the schools so that no two of them will be in the saine school?

Solution:

The number of ways of admitting four sons into five schools if no two of them will be in the same school.

\({ }^n P_r={ }^5 P_4\) = 5 . 4 . 3 . 2 = 120.

Question 25.

If there are 25 railway stations on a railway line, how many types of single second class tickets must be printed, so as to enable a passenger to travel from one station to another?

Solution:

Number of stations on a railway line = 25

Num ber of single different second class tickets must be printed so as to enable a passenger to travel from one station to another = Number of ways of arranging two station names out of 25 station names = \({ }^{25} \mathrm{P}_2\)

= 25 . 24 = 600.

Question 26.

In a class, there are 30 students on the new year day, every student posts a greeting card to all his/her classmates. Find the total number of greeting cards posted by them.

Solution:

Total number of students in a class = 30.

Total number of greeting cards posted by all the students to their classmates = number of ways of arranging names of two students from 30 names of students

= \({ }^n P_r={ }^{30} P_2\)

= 30 × 29 = 870.

Question 27.

Find the number of 4 letter words that can be formed using the letters of the word PISTON in which atleast one letter is repeated. [AP – Mar. 2015]

Solution:

The given word has six letters.

When repetition is allowed:

The number of four letter words that can be formed using these six letters when repetition is allowed is \(n^r=6^4\) = 6 . 6 . 6 . 6 = 1296.

When repetition is not allowed:

The number of four letter words that can be formed using these letters of the six letters when repitition is not allowed is

\({ }^n \mathrm{P}_{\mathrm{r}}={ }^6 \mathrm{P}_4\) = 360

The number of four letter words in which atleast one letter repeated is \(n^r-{ }^n P_r=6^4-{ }^6 P_4\) = 1296 – 360 = 936.

Question 28.

A number lock has 3 rings and each ring has 9 digits 1, 2, 3, ……………, 9. Find the maximum number of unsuccessful attempts that can be made by a persoti who tries to open the lock without knowing the key code.

Solution:

Each ring can be rotated in 9 different ways.

The total no. of different ways in which three kings can be rotated is = 9^{3} = 9 . 9 . 9 = 729

Out of these attempts, only one attempt is successful attempt.

Therefore, the maximum no. of unsuccessful attempts is 729 – 1 = 728.

Question 29.

Find the number of functions from a set A containing 5 elements into a set B containing 4 elements.

Solution:

Given number of A = n(A) = m = 5

number of B = n(B) = n = 4

Set A contains 5 elements and set B contains 4 elements.

The total number of functions from set A containing m elements to set B containing n elements is \(n^m=4^5\) = 1024.

Question 30.

Find the number of bijections from a set A contaIning 7 elements onto itself.

Solution:

Given n(A) = n = 7.

∴ The number of bijections from set A with n elements to set B with same number of elements ‘n’, A is n!.

The number of bijections from set A with 7 elements onto itself = 7!

= 7. 6 . 5 . 4 . 3 . 2 . 1 = 5040.

Question 31.

Find the number of ways of arranging 7 persons around a circle.

Solution:

Given number of persons, n = 7.

∴ The number of ways of arranging 7 persons around a circle is

(n – 1)! = (7 – 1)! = 6!

= 6 . 5 . 4 . 3 . 2 . 1 = 720.

Question 32.

Find the number of ways of preparing a chain with 6 different coloured beads. [TS – Mar. ‘19, 16; March ‘08]

Solution:

Neglecting the directions of beads in the chain.

Number of ways of preparing a chain with 6 different coloured beads = \(\frac{(n-1) !}{2}\)

= \(\frac{(6-1) !}{2}=\frac{5 !}{2}=\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2}\) = 60.

Question 33.

Find the number of ways of arranging the Chief Minister and 10 Cabinet Ministers at a circular table so that the Chief Minister always sits in a particular seat.

Solution:

Total number of persons = 11

Chief Minister can sit in a particular seat in one way.

Now, remaining positions are well defined relative to Chief Minister.

Hence, the remaining can sit in 10 places in 10! ways.

∴ The number of required arrangements = 10! × 1

= 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1 = 3628800.

Question 34.

Find the number of ways of arranging the letters of the word a^{4} b^{3} c^{5} in its expanded form.

Solution:

The expanded form of a^{4} b^{3} c^{5} is a . a . a . a . b . b . b . c . c . c . c . c.

This word has 12 letters in which there are 4a’s, 3b’s, 5c’s.

∴ They can be arranged in \(\frac{12 !}{4 ! 3 ! 5 !}\) ways.

Question 35.

There are 4 copies (alike) each of 3 different books. Find the number of ways of arranging these 12 books in a shelf in a single row.

Solution:

We have 12 books in which 4 books are alike of 1 kind. 4 books are alike of 2nd kind and 4 books are alike of 3rd kind.

Hence, they can be arranged in a shelf in a row in \(\frac{12 !}{4 ! 4 ! 4 !}\) ways = \(\frac{12 !}{(4 !)^3}\) ways.

Question 36.

Find the number of ways of arranging the letters of the word INDEPENDENCE. [TS – May 2016; March ’09, May ’13].

Solution:

Given word is INDEPENDENCE

There are 12 letters in INDEPENDENCE, in which there are 3 Ns are alike. 2 D’s are alike, 4 E’s are alike, and rest are different.

∴ The no. of required arrangements = \(\frac{12 !}{3 ! \cdot 2 ! \cdot 4 !}\)

Question 37.

Find the number of ways of arranging the letters of the word MATHEMATICS. [AP & TS – Mar. ’18; May ’97, ’10, March ’11, ’06]

Solution:

Given word is MATHEMATICS.

The word MATHEMATICS contains 11 letters in which there are , 2 M’s are alike, 2 A’s are alike, 2 T’s are alii ice and rest are different.

∴ The no. of required arran igements 11! = \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\).

Question 38.

Find the number of ways of arranging the letters of the word INTERMEDIATE. [AP -Mar. ’19; May 2016; May ’14, Board Paper]

Solution:

Given word is INTERMEDIATE.

The word INTERMEDIATE contains 121 letters in which there are 2 I’s are alike, 2 T’s are alike, 3 E’s are alike and rest are different.

∴ The no. of required arrangements = \(\frac{12 !}{2 ! \cdot 2 ! \cdot 3 !}\).

Question 39.

Find the number of 7 digit numbers that can be fonned using 2, 2, 2, 3, 3, 4 4.

Solution:

In the given seven digits, there are three 2’s, two 3’s, two 4’s.

∴ The number of seven digited numbers that can be formed using th given digits = \(\frac{7 !}{3 ! \cdot 2 ! \cdot 2 !}\).

Question 40.

Find the number of ways of selecting 7 members from a continent of 10 soldiers.

Solution:

The number of ways of selecting 7 members out of 10 soldiers is \({ }^{10} \mathrm{C}_7\) = 120.

Question 41.

A set A has 8 elements, find the number of subsets of A, containing atleaist 6 elements.

Solution:

The no. of subsets of A containing atlast 6 elements then the number of subsets of A containing 6 or 7 or 8 elements.

Number of subsets of A containing exactly 6 elements = \({ }^8 \mathrm{C}_6\)

Number cf subsets of A containing exactly 7 elements = \({ }^8 \mathrm{C}_7\)

Number of :ubsets of A containing exactly 8 elements = \({ }^8 \mathrm{C}_8\)

The nurrber of subsets of A containing at least 6 elements = \({ }^8 \mathrm{C}_6+{ }^8 \mathrm{C}_7+{ }^8 \mathrm{C}_8\)

= \({ }^8 \mathrm{C}_2+{ }^8 \mathrm{C}_1+{ }^8 \mathrm{C}_0\)

= \(\frac{8 \cdot 7}{2 \cdot 1}\) + 8 + 1

= 28 + 8 – 1 = 37.

Question 42.

Find the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls. [TS – Mar. 2015]

Solution:

4 boys can be selected from the given boys in \({ }^8 \mathrm{C}_4\) ways.

3 girls can be seleted from the given 5 girls in \({ }^5 \mathrm{C}_3\) ways.

∴ The required number of selections is \({ }^8 \mathrm{C}_4 \cdot{ }^{! 5} \mathrm{C}_3\) = \(\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} \frac{5 \cdot 4}{2 \cdot 1}\) = 700.

Question 43.

If there are 5 alIke pens, 6 alike pencils and 7 alike erasers, find the number of ways of selecting any number of (one or

more ) things out of them.

Solution:

The required number of ways is (p + 1) (q + 1) (r + 1) – 1

= (5 + 1) (6 + 1) (7 + 1) – 1

= 6 . 7 . 8 – 1

= 336 – 1 = 335.

Question 44.

To pass an examination a student has to pass in each of the three papers. In how many ways can a student fail in the exaimination? [TS- May 2015]

Solution:

For each of the three papers there are two choices P or F.

There are 2^{3} = 8 choices.

But a student passes only il he/she passes in all papers.

∴ Required number of ways = 2^{3} – 1 = 7.

Question 45.

In a class, there are 30 students. If each student plays a chess game with each of the other student, then find the total number of chess games played by them.

Solution:

Number of students in a class is 30.

Since each student plays a chess game with each of the student the total number of games played by them = \({ }^{30} \mathrm{C}_2\)

= \(\frac{30 \cdot 29}{2}\) = 435.

Question 46.

Find the number of diagonals of a polygon with 12 sides. [AP – May 2015]

Solution:

Number of sides of a polygon = 12

Number of diagonals of a n – sided polygon = \({ }^n C_2\) – n

∴ Number of diagonals of 12 sided polygon = \({ }^{12} \mathrm{C}_2\) – 12 = 54.

Question 47.

If \({ }^{\mathrm{n}} \mathrm{P}_3\) = 1320, find n. [May ‘08, March ‘05]

Solution:

12

Question 48.

If \({ }^{(n+1)} P_5:{ }^n P_6\) = 2 : 7, find n. [March ‘10, ‘07]

Solution:

11

Question 49.

If \({ }^{18} P_{(r-1)}:^{17} P_{(r-1)}\) = 9 : 7, find ‘r’.

Solution:

5

Question 50.

FInd the number of different chains that can be prepared using 7 different coloured beads. [AP – Mar. 2017]

Solution:

360