{"id":4602,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-6\/"},"modified":"2016-05-17T11:02:23","modified_gmt":"2016-05-17T05:32:23","slug":"ncert-solutions-class-11-maths-miscellaneous-6","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-6\/","title":{"rendered":"NCERT Solutions class-11 Maths Miscellaneous"},"content":{"rendered":"

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

1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?<\/strong><\/p>\n

Ans. <\/strong>There are 8 letters in the word DAUGHTER. In this word 3 vowels and 5 consonants. 2 vowels and 3 consonants are to be selected.<\/p>\n

Number of ways of selection = = = = 30<\/p>\n

Now, each word contains 5 letters which can be arranged among themselves 5! Ways.<\/p>\n

Therefore, total number of words = = 3600<\/p>\n

\n

2. How many words, with or without meaning can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?<\/strong><\/p>\n

Ans. <\/strong>There are 8 letters in the word EQUATION. In this word 5 vowels and 3 consonants. 2 vowels and 3 consonants are to be selected.<\/p>\n

Now, 5 vowels can be arranged in 5! Ways and 3 consonants can be arranged in 3! Ways.<\/p>\n

Also the groups of vowels and consonants can be arranged in 2! Ways.<\/p>\n

Total number of permutations = = = 1440<\/p>\n

\n

3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:<\/strong><\/p>\n

(i) exactly 3 girls <\/strong><\/p>\n

(ii) at least 3 girls <\/strong><\/p>\n

(iii) almost 3 girls?<\/strong><\/p>\n

Ans. (i)<\/strong> There are 9 boys and 4 girls. 3 girls and 4 boys have to be selected.<\/p>\n

Number of ways of selection = = <\/p>\n

= = 504<\/p>\n

(ii)<\/strong> We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.<\/p>\n

Number of ways of selection = <\/p>\n

= <\/p>\n

= <\/p>\n

= 504 + 84 = 588<\/p>\n

(iii)<\/strong> We have to select at most 3 girls. So the committee consists of no girls and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.<\/p>\n

Number of ways of selection = <\/p>\n

= <\/p>\n

= <\/p>\n

<\/p>\n

= 36 + 336 + 756 + 504 = 1632<\/p>\n

\n

4. If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?<\/strong><\/p>\n

Ans. <\/strong>In the word EXAMINATION, there are two I\u2019s and two N\u2019s and all other letters are different.<\/p>\n

Number of ways of arrangement = <\/p>\n

= <\/p>\n

= 907200<\/p>\n

\n

5. How many 6-digit numbers can be formed from the digits 0, 1 3, 5, 7 and 9 which are divisible by 10 and no digits are repeated?<\/strong><\/p>\n

Ans. <\/strong>A number divisible by 10 have unit place digit 0. So digit 0 is fixed at unit place and the remaining 5 placed filled with remaining five digits in ways.<\/p>\n

Required numbers = = = 120<\/p>\n

\n

6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabets?<\/strong><\/p>\n

Ans. <\/strong>2 vowels out of 5 vowels and 2 consonants out of 21 consonants have to be selected and these 4 letters in 4 ways.<\/p>\n

Required number of words = = <\/p>\n

= = = 50400<\/p>\n

\n

7. In an examination, a question paper consists of 12 questions divided into two parts i.e., part I and part II containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?<\/strong><\/p>\n

Ans. <\/strong>Here, we have to select 8 questions at least 3 questions from each section. Therefore, we have required selections are 3 from part I and 5 from part II or 4 from part I and 4 from part II or 5 from part I and 3 from part II.<\/p>\n

Number of ways of selection = <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= 210 + 175 + 35 = 420<\/p>\n

\n

8. Determine the number of 5-card combinations out of a deck of 52 cards is each selection of 5 cards has exactly one king.<\/strong><\/p>\n

Ans. <\/strong>Here, we have to select 5 cards containing 1 king and 4 other cards i.e., we have to select 1 king out of 4 kings and 4 other cards out of 48 other cards.<\/p>\n

Number of ways of selection = <\/p>\n

= <\/p>\n

= <\/p>\n

= = 778320<\/p>\n

\n

9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?<\/strong><\/p>\n

Ans. <\/strong>Given: women occupy the even places.<\/p>\n

1 2 3 4 5 6 7 8 9<\/p>\n

M W M W M W M W M<\/p>\n

Therefore, we can arrange four women in 4! Ways and 5 men in 5! Ways.<\/p>\n

Number of ways of selection = = 2880<\/p>\n

\n

10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?<\/strong><\/p>\n

Ans. <\/strong>According to the question, Number of ways of selection = <\/p>\n

= <\/p>\n

= <\/p>\n

= 170544 + 646646 = 817190<\/p>\n

\n

11. In how many ways can be letters of the word ASSASSINATION be arranged so that all the S\u2019s are together?<\/strong><\/p>\n

Ans. <\/strong>In the word ASSASSINATION, A appears 3 times, S appears 4 times, I appears in 2 times and N appears in 2 times. Now, 4 S\u2019 taken together become a single letter and other remaining letters taken with this single letter.<\/p>\n

Number of arrangements = <\/p>\n

= 151200<\/p>\n","protected":false},"excerpt":{"rendered":"

Miscellaneous Exercise 1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? Ans. There are 8 letters in the word DAUGHTER. In this word 3 vowels and 5 consonants. 2 vowels and 3 consonants are to be selected. Number of … 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-4602","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Miscellaneous | 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-miscellaneous-6\/\" \/>\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 Miscellaneous | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Miscellaneous Exercise 1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? Ans. There are 8 letters in the word DAUGHTER. In this word 3 vowels and 5 consonants. 2 vowels and 3 consonants are to be selected. Number of ... 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