{"id":4600,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-7-3\/"},"modified":"2016-05-17T11:01:02","modified_gmt":"2016-05-17T05:31:02","slug":"ncert-solutions-class-11-maths-exercise-7-3","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-7-3\/","title":{"rendered":"NCERT Solutions class-11 Maths Exercise 7.3"},"content":{"rendered":"

Exercise 7.3<\/strong><\/p>\n

1.<\/strong> How many 3-digit numbers can be formed by using the digits 1 to 9 if no digits is repeated?<\/strong><\/p>\n

Ans. <\/strong>Total number of digits = 9 and Number of digits used without repetition = 3<\/p>\n

Number of permutations = = = 504<\/p>\n

\n

2. How many 4-digit numbers are there with no digit repeated?<\/strong><\/p>\n

Ans. <\/strong>Total number of digits = 10 and Number of digits used without repetition = 4<\/p>\n

0 cannot be filled in the fourth place therefore number of permutations for fourth place = 9<\/p>\n

Number of permutations = = = 504<\/p>\n

Hence total number of permutations = = 4536<\/p>\n

\n

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?<\/strong><\/p>\n

Ans. <\/strong>Total number of digits = 6 and unit place can be filled with any one of the digits 2, 4, 6.<\/p>\n

Number of permutations = = = 3<\/p>\n

Now, the tens and hundreds places can be filled by remaining 5 digits.<\/p>\n

Number of permutations = = = 20<\/p>\n

Therefore, total number of permutations = = 60<\/p>\n

\n

4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?<\/strong><\/p>\n

Ans. <\/strong>Total number of digits = 5 and Number of digits used without repetition = 4<\/p>\n

Number of permutations = = = 120<\/p>\n

Now, unit place can be filled with any one of the digits 2, 4 for even number.<\/p>\n

Number of permutations = = = 2<\/p>\n

Now, the remaining three places can be filled with the remaining 4 digits.<\/p>\n

Number of permutations = = = 24<\/p>\n

Therefore, total number of permutations of even numbers = = 48<\/p>\n

\n

5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?<\/strong><\/p>\n

Ans. <\/strong>Total number of persons = 8 and Number of persons used without repetition = 2<\/p>\n

Number of permutations = = = 56<\/p>\n

\n

6. Find if <\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

7. Find if: <\/strong><\/p>\n

(i) <\/strong><\/p>\n

(ii) <\/strong><\/p>\n

Ans. (i)<\/strong> Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

or <\/p>\n

is not possible because , therefore, <\/p>\n

(ii)<\/strong> Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

or <\/p>\n

is not possible because , therefore, <\/p>\n

\n

8. How many words, with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once?<\/strong><\/p>\n

Ans. <\/strong>Total number of letters in word EQUATION = 8<\/p>\n

Number of letters used (all different) = 8<\/p>\n

Number of permutations = = = 40320<\/p>\n

\n

9. How many words, with or without meaning can be formed using all the letters of the word MONDAY, assuming that no letter is repeated if:<\/strong><\/p>\n

(i) 4 letters are used at a time <\/strong><\/p>\n

(ii) all letters are used at a time<\/strong><\/p>\n

(iii) all letters are used by first letter is a vowel?<\/strong><\/p>\n

Ans. <\/strong>Total number of letters in word MONDAY = 6 and number of vowels = 2<\/p>\n

(i)<\/strong> Number of letters used = 4<\/p>\n

Number of permutations = = = 360<\/p>\n

(ii)<\/strong> Number of letters used = 6<\/p>\n

Number of permutations = = = 720<\/p>\n

(iii)<\/strong> First letter is vowel.<\/p>\n

Number of permutations of vowels = = = 2<\/p>\n

Now, remaining five places can be filled with remaining five letters.<\/p>\n

Number of permutations = = = 120<\/p>\n

Therefore, total number of permutations = = 240<\/p>\n

\n

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I\u2019s not come together?<\/strong><\/p>\n

Ans. <\/strong>Total letters of the word MISSISSIPPI = 11<\/p>\n

Here M = 1, I = 4, S = 4 and P =2<\/p>\n

Number of permutations = = 34650<\/p>\n

When the four I\u2019s come together then it becomes one letter so total number of letters in the word when all I\u2019s come together = 8<\/p>\n

Number of permutations = = 840<\/p>\n

Number of permutations when four I\u2019s do not come together = 34650 \u2013 840 = 33810<\/p>\n

\n

11. In how many ways can the letters of the word PERMUTATIONS be arranged if the:<\/strong><\/p>\n

(i) words start with P and end with S<\/strong><\/p>\n

(ii) vowels are all together<\/strong><\/p>\n

(iii) There are always 4 letters between P and S?<\/strong><\/p>\n

Ans. <\/strong>Total numbers in the word PERMUTATIONS = 12 and here T = 2<\/p>\n

(i)<\/strong> First letter is P and last letter is S which are fixed.<\/p>\n

Therefore, remaining 10 letters are to be arranged between P and S.<\/p>\n

Number of permutations = = 1814400<\/p>\n

(ii)<\/strong> There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.<\/p>\n

Number of permutations of all vowels together = = = 120<\/p>\n

Now consider the 5 vowels together as one letter. SO the number of letters in the word when all vowels are together = 8<\/p>\n

Number of permutations = = 20160<\/p>\n

(iii)<\/strong> P and S are on 1st<\/sup> and 6th<\/sup> places P and S are on 2nd<\/sup> and 7th<\/sup> places<\/p>\n

P and S are on 3rd<\/sup> and 8th<\/sup> places P and S are on 4th<\/sup> and 9th<\/sup> places<\/p>\n

P and S are on 5th<\/sup> and 10th<\/sup> places P and S are on 6th<\/sup> and 11th<\/sup> places<\/p>\n

P and S are on 7th<\/sup> and 12th<\/sup> places<\/p>\n

Now, P and S can be put in 7 ways and also P and S can interchange their positions.<\/p>\n

Number of permutations = = 14<\/p>\n

Now the remaining 10 places can be filled with remaining 10 letters.<\/p>\n

Number of permutations = = 1814400<\/p>\n

Therefore, total number of permutations = = 25401600<\/p>\n","protected":false},"excerpt":{"rendered":"

Exercise 7.3 1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digits is repeated? Ans. Total number of digits = 9 and Number of digits used without repetition = 3 Number of permutations = = = 504 2. How many 4-digit numbers are there with no digit … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47,281],"tags":[],"class_list":["post-4600","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Exercise 7.3 | myCBSEguide<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-7-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions class-11 Maths Exercise 7.3 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Exercise 7.3 1. 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