Free PDF For NCERT Maths Solutions- Fullmarkseducare.com

1. FULLMARKS EDUCARE (Discover Yourself) Permutations and Combinations CLASS­XI Maths Q 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there? Sol. When a coin is tossed, the number of possible outcomes is 2 i.e. head OR tail. So in three successive throws, the number of possible outcomes is 2 × 2 × 2 = 8 Exercise 7.1 Q 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed? Sol. (i) Repetition of the digits is allowed: The unit’s place can be filled with any of the given five digits i.e. in 5 ways. As digits can be repeated, therefore, ten’s and hundred’s places also can be filled with any of the given five digits. Thus, by the multiplication principle, the number of three- digit numbers that can be formed with the given digits is 5 × 5 × 5 = 125 (ii) Repetition of digits not allowed: If unit’s place is filled in first, then it can be filled with any of the given five digits i.e. in 5 ways. Then, the ten’s place can be filled with any of the remaining four digits, and the hundred’s place can be filled with any of the remaining three digits. Thus, by the multiplication principle, the number of three- digit numbers that can be formed without repeating the given digits is 5 × 4 × 3 = 60 Q 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other? Sol. The first colour in the signal flag can be chosen in 5 different ways and the second in 4 different ways. Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20. Exercise 7.2 Q 1: Evaluate Sol. (i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320 (ii) 4! = 1 × 2 × 3 × 4 = 24 3! = 1 × 2 × 3 = 6 ? 4! – 3! = 24 – 6 = 18 (i) 8! (ii) 4! – 3! Q 2: Is 3! + 4! = 7!? Sol. 3! = 1 × 2 × 3 = 6 4! = 1 × 2 × 3 × 4 = 24 3! + 4! = 6 + 24 = 30 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040 ? 3! + 4!? 7! Q 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Sol. The unit’s place can be filled only with 2 or 4 or 6 only i.e. in 3 ways. As the digits can be repeated, the ten’s place can be filled with any of the 6 digits i.e. in 6 different ways. Similarly, the hundred’s place also can be filled in 6 ways. Therefore, by multiplication principle, the required number of three digit even numbers is 6 × 6×3 = 108 8 ! ? ! Q 3: Compute 6 2 ! ? ? ? 8 ! ? ! 8 7 6 6 ! ? ? ! 8 7 2 ? ? ? 28 Sol. 6 2 ! 2 1 Q 3: How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated? Sol. The first place in the code can be filled with any of the first 10 letters i.e. in 10 ways. As the digits cannot be repeated, the second place can be filled in 9 ways. Similarly, the third and the fourth places can be filled in 8 and 7 ways respectively. Thus the total number of the required codes is 10 × 9 × 8 × 7 = 5040 Q 4: If 1 1 7 x ? ? , find x. 6 ! ! 8 ! 1 6 1 7 x ? ? Sol. Given: ! ! 8 ! 1 6 1 ? x ? ? ? 8 7 6 ? ? ! 7 6 ! ! ? ? ? ? ? ? 1 6 1 7 x ? ? ? 1 Q 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once? Sol. As each telephone number starts with 67, we have only three places 67, _, _, _to fill. As digits cannot be repeated, we have to exclude 6 and 7. So, the first place can be filled in eight ways i.e. by any of the digits from 0 to 9, except digits 6 and 7. Similarly, the second place can be filled in 7 ways and the third in 6 ways. Therefore, the number of ways in which 5-digit telephone numbers can be constructed is 8 × 7 × 6 = 336 8 7 6 ? ? ! ! 1 7 x ? ? ? 1 8 7 ? 8 7 x ? ? 8 7 ? ? ? 8 8 7 7 ? ? x ? ? x 64 NCERTSOLUTIONS www.fullmarkseducare.com Ph:.9911968787

2. Q 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? Sol. Given digits: 1, 2, 3, 4, and 5. To form: 4-digit numbers with the digits. There are as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time. Therefore, required number of 4 digit numbers n! ? Q 5: Evaluate , when n r ! (i) n = 6, r = 2 Sol. (i) When n = 6, r = 2: (ii) n = 9, r = 5 ? ? n! ? 6 ? ! 6 4 ! ! 6 5 4 4 ! ? ? ? ? 30 n r ! 6 2 ! ! (ii) When n = 9, r = 5: 5 ? ! 5 1 ! ! ? ? ? 5 P 4 ? ? ? ? n! ? 9 ? ! 9 4 ! ! 9 8 7 6 4 4 ! 5 4 ! ? ? ? n r ! 9 5 ! ! 1 2 3 4 5 ? ? ? ? ? ? 120 9 8 7 6 5 ? ? ? ? ? ? 15120 Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4. The number of ways in which units place is filled with digits is 2. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits. Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time. Thus, number of ways to fill the remaining places Exercise 7.3 Q 1: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Sol. Here no digit is repeated. Also, the order of the digits matters. Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time. Therefore, required number of 3-digit numbers 4 ? ! 4 1 ! ! ? ? ? 4 P 9 ? ! 9 6 ! ! ? ? ? 9 8 7 6 6 ! ? ? ? 9 3 P 4 3 ! ? 9 8 7 ? ? ? ? 504 3 9 3 ! ! = 4 × 3 × 2 × 1 = 24 Q 2:How many 4-digit numbers are there with no digit repeated? Sol. The digit 0 cannot occupy the thousand’s place in the required number. Hence, the thousand’s place can be filled in 9 ways i.e.with any of the digits from 1 to 9. However, the hundred’s, ten’s, and unit’s place can be filled with any of the digits from 0 to 9. As the digits cannot be repeated in this case, therefore the digit occupying the thousand’s place is excluded. At the same time the digit 0 which could not occupy the thousand’s place is available to fill the other places. Thus, 9 digits are available to fill the hundreds, tens, and units places. This can be done in 9 Thus, by multiplication principle, the required number of even numbers is 24 × 2 = 48 Q 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? Sol. Given: A committee of 8 persons., To choose: A chairman and a vice-chairman given that one person cannot hold more than one position. So, we have to find, the permutations of 8 different objects taken 2 at a time. Thus, required number of ways 3P ? ? 8 ? ! 8 6 ! ! 8 7 6 6 ! ? ? ? 8 7 ? ? ? 8 P 56 = 2 9 ? ! 9 6 ! != 504 8 2 ! ! ? ? Now, 9 P 3 9 3 ! ? ? n 1 n Q 6: Find n if Sol. P : P 1 9 : . 3 4 Thus, the required number of 4-digit numbers is 9 × 504 = 4536 Q 3: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated? Sol.Given digits: 1, 2, 3, 4, 6, and 7. To form: 3-digit even numbers without repeating the digits. Clearly, the unit’s place can be filled in 3 ways i.e. with any of the digits, 2, 4, or 6. As repetition of digits is not allowed, the digit occupying the unit’s place cannot fill hundred’s or ten’s place. Thus, only 5 digits are available for filling the hundred’s and ten’s place. So, the hundred’s and ten’s place can be filled with the remaining 5 digits in5 ? ? n 1 n P : P 1 9 : . 3 4 ? ? ? ? ? ? ? n 1 ! 1 3 ? ? n ! ? n 1 1 9 P P 1 9 ? ? ? ? 3 ? ? ? ? ? ? n n! 4 ? n 4 ! ? ? ? n n 1 4 ! ! n 4 ! 1 9 ? ? ? n! ? n 1 ? ! 2 P ways. 1 9 ? ? ? n n 1 ! 5! ? 5! 3! ? ? 5 P = 20 2 5 2 ! 1 n 1 9 ? ? ? ? n 9 Thus, by multiplication principle, the number of required 3- digit numbers is 3 × 20 = 60 NCERTSOLUTIONS www.fullmarkseducare.com Ph:.9911968787

3. ? ? Q 7: Find r if (i) 5 6 (ii)5 6 P 2 P? P P? 6 ? ! 6 2 ! != 360 r r 1 r r 1 ? ? 6 P 4 6 4 ! ? Sol. (i) Given: 5 6 P 2 P? r r 1 (ii) All letters used at a time: The number of such words is the number of permutations of 6 different objects taken 6 at a time, which is 6 6 6 P ! ? = 720. (iii) The given word has 2 different vowels. The first letter has to be a vowel, which can be chosen in 2 ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. This can be done in 5! ways. Thus, in this case, required number of words that can be formed is 5! × 2 = 120 × 2 = 240 ? 5 ? ! r ! 6 r ! 5 ? ! 2 6 7 ? ! ? ? ? ? ? 2 ? ? 5 6 1 ! 5 r ! r ! ? ? 5! ? 2 6 5! 6 ? ? ? ? ? 5 r ! 7 r r 5 r ! ? 2 6 r ? ? 1 ? ? 7 6 r ? ? ? ? ? ? ? ? 2 r 3 r 10 0 r 13 r 30 0 ? ? ? r 3 or r 10 Q 10:In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? Sol. In the word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the 11 letters in the given word n! ? ? We know that, n P , ? ? where,0 r n r n r ! Hence r ? 10, (ii) 5 r P ? ? r = 3 6 P? r 1 5 ? ! r ! 6 r ! ? ? ? ? ? ? ? 11 ! ! ! ! 11 10 9 8 7 6 5 4 4 4 3 2 1 2 1 ! ? ? ? ? ? ? ! ? ? ? ? = 34650 ? ? 5 6 1 ! 4 4 2 When the 4 I’s occur together, they can be treated as a single object. Thus, the total number of objects to be arranged becomes 8 with 4 S’s and 2 P’s. These objects ? 5 ? ! r ! 6 5 7 ? ! ? ? 5 r ! 8 ! !ways i.e., 840 ways. ! 1 ? 6 ? ? ? can be arranged in 4 2 ? ? 5 r ! 7 r 6 r 5 r ! Number of arrangements where all I’s occur together = 840 Thus, number of distinct permutations of the letters in MISSISSIPPI in which four I’s do not come together = 34650 – 840 = 33810 Q 11: In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S? Sol. The word PERMUTATIONS has 12 letters including 2 T’s. (i) When the words start with P and end with S, the remaining 10 places can be filled with the remaining 10 letters. As these letters include 2 T’s, the number of required permutations is ? ? ? ? 7 r 6 r 6 ? ? ? ? ? ? 2 r 13 r 36 ? 0 ? r 4 r 9 0 ? ? ? r 4 or r 9 n! ? ? ? ? We know that, n P where,0 r n . r n r ! Thus, 0 ? r ? 5 Hence, r ? 9 ? r = 4 Q 8: How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once? Sol. The word EQUATION has 8 different letters. Therefore, the number of words that can be formed using all the letters of the word EQUATION, without repetition, is the number of permutations of 8 different objects taken 8 at a time, which is 8 8 8 P ! ? 10 2 ! ? ? 1814400 ! (ii) The word has 5 vowels i.e. A, E, I, O, U. As the vowels occur together, they can be treated as a single object. This object along with the remaining 7 objects will account for 8 objects. These 8 objects in which there are 2 T’s can be arranged in = 40320 8 2 ! !ways. Q 9: How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel? Sol. There word MONDAY has 6 different letters. (i) 4-letter words without repetition: It is the number of permutations of 6 different objects taken 4 at a time, which is6 Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways. Therefore, by multiplication principle, required number of 8 2 ! ! ? ? ? arrangements in this case 5 ! 2419200 (iii) Always 4 letters between P and S. Consider PERMUSTATION as one case of this kind and SERMUPTATION as another where there are always 4 4 P . NCERTSOLUTIONS www.fullmarkseducare.com Ph:.9911968787

4. letters between P and S. We have 7 allocations possible to fix P and S in that way PERMUSTATION, and 7 allocations possible to fix S and P in that way SERMUPTATION. So, basically there are 14 ways to fix P and S, whereas the remaining 10 letters with only T occurring twice may be arranged in 10! /2! number of ways. Hence, the total arrangements in which there are always 4 letters between P 3 girls can be selected from 4 girls in 4 Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected C ways. 3 5! 4! ? ? ? ? 5 4 C C 3 3 3! 2! 3!1! Q 5: 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. Sol. Given: 6 red balls, 5 white balls, and 5 blue balls. To select: 3 balls of each colour. Here, 3 balls can be selected from 6 red balls in6 10 2 ! ? ? ? and S is 14 25401600 ! Exercise 7.4 C ways. 3 ? n n nC Q 1: If C , find C 8 2 2 3 balls can be selected from 5 white balls in5 C ways. 3 ? n n Sol. We know that if C then n = x + y C x y 3 balls can be selected from 5 blue balls in5 Thus, by multiplication principle, required number of ways of selecting 9 balls C ways. 3 ? n n Given C . Therefore, n = 8 + 2 = 10 C 8 2 10 ! 10 2 8 ! ! ! ? ? ? ? ? n 10 C C 45 2 2 ? 2 10 2 ! ! 6 ! ! ! 5 ! ! ! 5 ! ! ! ? ? ? ? ? ? 6 5 5 C C C 3 3 3 3 3 3 2 3 2 Q 2: Determine n if ? ? ? ? ? ? ? 6 5 4 3 3 ! ? ? ? ! 5 4 3 3 ! ? ? ! 5 4 3 3 ! ? ? ! 2 n C C 11 1 2 n C C 12 1 ? ? ? ? 3 ? (i) (ii) 3 3 2 ? 2 1 2 1 n n 3 3 ? ? 20 10 10 2000 2 n C C 12 1 Q 6: Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination. Sol. A deck of 52 cards has 4 aces. We have to select 5 cards with exactly one ace. One ace can be selected in4 ? 3 Sol. (i) Given: n 3 ? 2 n ! n ? 3 ! n 3 ! 12 1 ? ? ? 3 2 ! 3 ! n! C ways and the remaining 4 1 ? ? ? ? 2 n 2 n 1 2 2 n 2 2 n 3 ! n n 3 ! 2 ? ? ? ? 12 cards can be selected out of the 48 cards in48 Thus, by multiplication principle, required number of 5 card combinations C ways. ? ? ? n 3 ! n n 1 n 3 ! 4 ? 2 n n 1 2 ? ? 3 ? 48 ! ! 4 ! ! ! ? ? ? ? 48 4 C C 4 1 4 44 ! 1 3 ? ? ? ? 2 n 1 3 n 2 ? ? n 5 Q 7: In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers? Sol. Given: 17 players out of which 5 are bowlers. To select: A team of 11 players with exactly 4 bowlers. Now, 4 bowlers can be selected in5 remaining 7 players can be selected out of the 12 players in12 7 C ways. Thus, the number of ways of selecting the cricket team ? 2 ! n n 3! n 3 ! 2 n C C 11 1 ? ? ? 11 ? 3 (ii) ? 3! 2 3 ! n ! n 3 ? ? ? ? 2 n 2 n 1 2 2 n 2 2 n 3 ! n n 3 ! 2 ? ? ? ? 11 ? ? ? n 3 ! n n 1 n 3 ! C ways and the 4 ? ? 2 2 n 1 2 1 n ? 2 ? 4 2 n ? 1 ? ? 11 ? ? 11 ? n n 2 n 2 ? Q 3: How many chords can be drawn through 21 points on a circle? Sol. A chord is a line that joins any two points on lying on the circumference of the circle. Therefore, the required number of chords ? ? ? ? 4 2 n 1 11 n 2 n = 6 ? ? ? ? 5 ! ! ! 12 7 5 ! ! ! 12 11 10 9 8 5 4 3 2 1 ? ? ? ? ? ? ? ? ? ? ? 5 12 C C 5 3960 4 7 4 1 Q 8: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected. Sol. 2 black balls can be selected out of 5 black balls in5 ways and 3 red balls can be selected out of 6 red balls in6 3 C ways. Thus, the number of ways of selecting 2 black and 3 red C ? 21! 21! 2!19! 21 20 2 2 ? ? ? ? 21 C 210 = 2 ? 2! 21 2 ! Q 4: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? ? ? ? 5 ! ! ! 6 ! ! ! 5 4 2 6 5 4 3 2 1 ? ? 5 Sol. 3 boys can be selected from 5 boys in C ways. ? ? ? ? ? ? 5 6 C C balls = 200 3 2 3 2 3 3 3 NCERTSOLUTIONS www.fullmarkseducare.com Ph:.9911968787

5. ? ? ? =4 9 4 9 = 588 C C C C 3 4 4 3 Q 9: In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student? Sol. Given: 9 courses out of which, 2 are compulsory. To select: A total of 5 courses. Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be done in7 (iii) At the most 3 girls: (a) 3 girls and 4 boys (c) 1 girl and 6 boys 3 girls and 4 boys can be selected in4 (b) 2 girls and 5 boys (d) No girl and 7 boys ? 4 ways. C C 3 4 ? 2 girls and 5 boys can be selected in4 9 ways. C C 2 5 C ways. 3 ? 1 girl and 6 boys can be selected in4 9 ways. C C 1 6 Thus, required number of ways = 7 C= 35 3 ? No girl and 7 boys can be selected in4 Therefore, the required number of ways ? ? ? ? C C 9 ways. C C 0 7 Miscellaneous Exercise on Chapter 7 ? ? ? ? 4 9 4 9 4 9 4 9 C C C C C C 3 4 2 5 1 6 0 7 4 ! ! ! 9 ! ! ! 4 ! ! ? ! 9 ! ! ! 4 ! ! ! 9 ! ! ! 4 ! ! ! 9 ! ! ! ? ? ? ? ? ? ? ? Q 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? Sol. The word DAUGHTER has 3 vowels (A, U, and E), and 5 consonants (D, G, H, T, and R). 2 vowels out of 3 vowels can be selected in3 3 1 4 5 2 2 5 4 ? 1 3 6 3 0 4 7 2 ? 504 756 336 36 ? ? 1632 Q 4: If the different permutations of all the letters of the word EXAMINATION are listed as in dictionary, how many words are there in this list before the first word starting with E? Sol. The word EXAMINATION has 11 letters out of which, A, I, and N appear two times and all the other letters appear only once. The words that appear before the first words starting with E will be the words that start with A. In these words, letter A occupies the first place on the left. The remaining letters are rearranged among the rest of the 10 places. The remaining ten letters include 2 I’s and 2 N’s. Thus, the C ? ways. 3 2 C ? 3 consonants out of 5 can be selected in5 Therefore, the total number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30 Letters in each of these 30 combinations can be arranged among themselves in 5! ways. Hence, required number of different words = 30 × 5! = 3600 ways. 10 3 Q 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? Sol. The word EQUATION has 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N). Required: All vowels and all consonants occur together. Thus the vowels (A, E, I, O, U) can be treated as one object. Similarly, the consonants (Q, T, N) can be treated as one objects. These 2 objects can be arranged in 2 The vowels among themselves can be arranged in 5! ways. Similarly, the 3 consonants can be arranged among themselves in 3! ways. Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440 number of words starting with A = 10! 2!2!? 907200 Thus, the required numbers of words is 907200. Q 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 digit is repeated? Sol. A number is divisible by 10 if its unit’s digit is 0. Therefore, 0 is fixed at the unit’s place. As repetition of the digits is NOT allowed, the remaining five places can be filled with the digits 1, 3, 5, 7 and 9 in 5! ways. Hence, the number of required 6-digit numbers = 5! = 120 ? 2 P 2! Q 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: (i) exactly 3 girls? (ii) atleast 3 girls? (iii) atmost 3 girls? Sol. The committee has 7 members to be selected from among 9 boys and 4girls. (i)Exactly 3 girls: Number of boys in each committee = (7 – 3) = 4. Thus, the required number of ways = Q 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 alphabet? Sol. We have to select 2 vowels and 2 consonants from the 5 vowels and 21 consonants of the English alphabet. 5 ! ! ! ? ? Two vowels can be selected in =5 C 10 ways 2 2 3 21 ! ! 4! 9! ? ? 21 ? ? ? C 210 4 9 C C Two consonants can be selected in = = 504 2 3 4 2 19 ! 3! 1! 4! 5! Therefore, the total number of combinations with 2 vowels and 2 consonants = 10 × 210 = 2100. The four letters in each of these 2100 combinations can be arranged among themselves in 4! ways. Therefore, required number of words = 2100 × 4! = 50400 (ii) At least 3 girls: Following are the possible compositions of the committee: (a) 3 girls and 4 boys or (b) 4 girls and 3 boys 3 girls and 4 boys can be selected in4 ? 9 ways. C C 3 4 ? 4 girls and 3 boys can be selected in4 Therefore, the required number of ways 9 ways. C C 4 3 Q 7: In an examination, a question paper consists of 12 NCERTSOLUTIONS www.fullmarkseducare.com Ph:.9911968787

6. 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? Sol. Part I contains 5 questions and Part II, 7 questions. Eight questions to be attempted with at least three from each part. This can be done in the following ways: (a) 3 Qs from part I and 5 Qs from part II (b) 4 Qs from part I and 4 Qs from part II (c) 5 Qs from part I and 3 Qs from part II 3 Qs from part I and 5 Qs from part II can be selected in ? ways. 4 Qs from part I and 4 Qs from part II can be selected in ? C C ways. 5 Qs from part I and 3 Qs from part II can be selected in5 5 3 C C ? ways. arrangements are possible? Sol. The 5 men can be seated in 5! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places). M M M M M ? ? ? ? Therefore, the women can be seated in 4! ways. Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 2880 Q 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? Sol. Case I: The three students join. Then, the remaining 7 students can be chosen from the remaining 22 students in 22 Case II: None of the three students joins. Then, 10 students can be chosen from the remaining 22 students in22 10 C ways. Thus, required number of ways of choosing the excursion party is22 7 10 ? C C . 5 7 C C 3 5 5 7 C ways. 4 4 7 7 Thus, the 8 questions can be selected in ? ? ? ? C C C ? ? 5 7 5 7 5 7 C ways. C C 3 5 4 4 5 3 22 5 ! ! ! 7 ! ! ? ! 5 ! ! ? ! 7 ! ! ! 5 ! ! ! 7 ! ! ! ? ? ? ? ? ? 2 3 2 5 4 1 4 3 5 0 3 4 Q 11: In how many ways, can the letters of the word ASSASSINATION be arranged so that all the S’s are together? Sol. In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once. All letters have to be arranged in such a way that the four S’s are together. Thus, we can treat SSSS as a single object. Thus total objects to be arranged = (13 -- 4) + 1 = 10. These 10 objects include 3 A’s, 2 I’s, and 2 N’s. These can ? 210 175 35 ? 420 Q 8: Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king. Sol.A deck of 52 cards, has 4 kings. One king can be selected out of 4 kings in 4 4 cards out of the remaining 48 cards can be selected in C ways. 1 48 C ways. Thus, the required number of 5-card combinations is ? . 10 ! ! ! ways. ! 4 be arranged in 3 2 2 4 48 C C Thus, the required number of ways of arranging the letters 1 4 10 ! ! ! ! ? ? of the given word Q 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 151200 3 2 2 NCERTSOLUTIONS www.fullmarkseducare.com Ph:.9911968787