Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Permutation and Combinations, Exercises of Mathematics

SRM Institute Of Science & TechnologyMathematics

Few practice problems on Permutation and Combinations

Typology: Exercises

2016/2017

Uploaded on 09/28/2017

harsh-anchalia
harsh-anchalia 🇮🇳

5

(1)

1 document

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Permutations
Problems:-
1) Find the value of 22p5
Sol: Formula npr = n(n-1)(n-2)(n-(r-1))
22P5 = 22(22-1)(22-2)…(22-(5-1))
=22(22-1)(22-2)(22-3)(22-4)
= 22(21)(20)(19)(18)
= 3160080
2) In how many ways 22 oranges can be arranged in 4 baskets?
Sol: Formula npr = n!/(n-r)!
22P4 = 22!/(22-4)!
= 22!/18!
= 175560
3) If np6=30.np4 find n
Sol: np6= 30.np4
n!/(n-6)! = 30. n!/(n-4)!
n!/(n-6)! = 30. n!/(n-4)(n-5)(n-6)!
(n-4)(n-5) = 30
n2-4n-5n+20 = 30
pf3
pf4
pf5

Partial preview of the text

Download Permutation and Combinations and more Exercises Mathematics in PDF only on Docsity!

Permutations

Problems:-

  1. Find the value of 22 p 5 Sol: Formula npr = n(n-1)(n-2)(n-(r-1)) (^22) P 5 = 22(22-1)(22-2)…(22-(5-1))

=22(22-1)(22-2)(22-3)(22-4) = 22(21)(20)(19)(18) = 3160080

  1. In how many ways 22 oranges can be arranged in 4 baskets? Sol: Formula npr = n!/(n-r)! (^22) P 4 = 22!/(22-4)!

= 22!/18! = 175560

  1. If np 6 =30.np 4 find n Sol: np 6 = 30.np 4 n!/(n-6)! = 30. n!/(n-4)! n!/(n-6)! = 30. n!/(n-4)(n-5)(n-6)! (n-4)(n-5) = 30 n^2 -4n-5n+20 = 30

n^2 -9n+20-30 = 0 n^2 -9n-10 = 0 n^2 +n-10n-10 = 0 n(n+1)-10(n+1) = 0 (n-10) (n+1) = 0 n = 10, - So n = 10 is the final value

  1. How many numbers lying between 150 and 600 can be formed with the help of digits 0, 1, 2, 3, 4? Sol: Here n=5 and r= So 5 p 3 = 5!/(5-3)! = 5!/2! = 120/2 = 60 But using the given five digits we can also form numbers like 012, 013, 014, 021, 023, 024, 031, 032, 034, 041, 042, 043, 102,103, 104, 120, 123, 124, 130, 132, 134, 140, 142, 143. These numbers doesn’t come under the category of 150 and 600. So these 24 numbers should be excluded. So 60-24 = 36 Hence number of digits lying between 150 and 600 that can be formed using digits 0, 1, 2, 3, 4 are 36.

  2. How many ways can be arranged with the help of 7pants and 6shirts such that no two pants are adjacent? Sol: First all the 7pants are arranged in 7! Ways So these 6shirts can be arranged in 6! Ways P 1 S 1 P 2 S 2 P 3 S 3 P 4 S 4 P 5 S 5 P 6 S 6 P 7

  1. Find the arrangements that can be made of the word ASSOCIATE?

Sol: Arrangements made of the word ASSOCIATE = 9!/2!2!1!1!1!1!1! = 90720

  1. How many different words can be formed with the letters of PSUCIDE. In how many of these P&D are together and how many of them begin with P and end with D? Sol: Number of words can be formed with the letters of PSUCIDE = 7! = 5040 P&D are together. So count them as one letter. Hence 6letters (S, U, C, I, E, P&D) can be arranged in 6! Ways. Those two letters P&D which are taken together can be rearranged in 2! Ways So P&D together are 6!2! = 7202 = 1440 Begin with P and end with D, so the remaining 5 letters can be arranged in 5! Ways So begin with P and end with D are 5! = 120

  2. The letters of the word ZEBRA are written in all possible orders. How many words are possible if these words are written in the dictionary order. What is the rank of ZEBRA? Sol: Number of words that are possible = 5! = 120 Total number of words begin with A is same as number of possible arrangements of B, E, R, Z. So total number of letters beginning with A is 24, B is 24, E is 24, R is 24, Z is 24. So words begin with Z is 97 to 120.

The rank of words beginning with ZA will be between 97 to 102.Similarly the rank of words beginning with ZE will be between 109 to 114.

The rank of words beginning with ZEA will be 109,110. Similarly the rank of words beginning with ZEB will be 111,112. The rank of words beginning with ZEB should be considered.

The rank of the word ZEBAR will be 111 and rank of the word ZEBRA will be

So rank of the word ZEBRA is 112.