PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1RgzGUTYnW4dhCa055vyUmFCRwA1gByc8/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true Permutations and Combinations Fundamental principle of counting; Permutation as an arrangement and combination as selection, Meaning of P(n,r) and C (n,r). Simple applications. Factorial-notation n or n! is read as ‘n factorial’ n = n! = 1.2.3……. (n – 1) n. m! = n! when m = n 1! = 1; ⎣0 = 1 Note : Factorial of negative numbers or proper fractions is not defined. FUNDAMENTAL PRINCIPLE OF COUNTING Multiplication theorem If an operation can be performed in ‘m’ ways and another operation can be performed in ‘n’ ways, then the two operations in succession can be performed in ‘m × n’ ways. For example: A person can go by rail from Delhi to Lucknow in two ways and from Lucknow to Varanasi in three ways then he can travel from Delhi to Varanasi via Lucknow in 2 × 3 = 6 ways Addition Theorem If an operation can be performed in ‘m’ ways and another independent operation can be performed in ‘n’ ways, then either of the two operations can be performed in ‘m + n’ ways. PERMUTATION AS AN ARRANGEMENT Each of the arrangements which can be made by taking some or all of a number of things is called a permutation. Let n and r be non-negative integers such that 0 ≤ r ≤ n. Then Number of permutation of n different things, taken r at a time, is denoted by nPr or P(n, r). C H A P T E R CHAPTER INCLUDES : Fundamental principle of counting Permutation as an arrangement Permutations under restrictions Circular permutations Combination as selection Key results on nc r Simple applications of combinations Derangements Solved examples n Pr n! (n−r )! = n(n − 1)(n − 2) (n − r + 1) In how many different sequences can seven questions be answered by an examinee out of 12 questions in the question paper? Solution : Total number of arrangements = 12P7 = 12! 5! = 12 × 11 × 10 × 9 × 8 × 7 × 6 = 3991680 Number of permutations of n different things taken all at a time is nPn = n! TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 In how many total ways, the letters of word “REPOST” can be arranged? Solution : Since there are 6 different letters, hence total number of arrangements = 6P6 = 6! = 720. Number of permutations of n things, taken all at a time, out of which p things are alike and of one n! type, q things are alike of second type and rest are all different is p!q! In how many more ways can the letters of “Aakash Institute” can be arranged? Solutions : Total there are 15 letters aaa , t t t , i i , ss , k, h, n, u, e 3 3 2 2 5 Hence total number of permutation = 15! 3!3!2!2! = 908, 107, 2000 Hence additional ways = 908,107,1999 The number of permutations of n different things, taken r at a time, when each thing can be repeated any number of times is nr. In how many ways can different four digit numbers be made by using the digits 1, 2, 3, 4, 5, 6? Solution : Since each place can be filled in six ways. Hence total number of numbers = 64 = 1296. PERMUTATIONS UNDER RESTRICTIONS Number of permutations of n different things taken r at a time, when one particular thing is to be always included is, r.n–1Pr–1. Number of permutation of n different things taken r at a time, when p particular things are to be always included, is r! (r − p)! ⋅n −p Pr −p Number of permutations of n different things taken r at a time, when p particular things are always together is p!.[r − (p − 1)] .n−p Pr −p Number of permutations of n different things, taken r at a time, when p particular things are not to be taken is n–pPr . Number of permutations of n different things, taken all at a time, when p particular things always occur together is p!(n – p + 1)! Number of permutations of n different things, taken all at a time, when m particular things never occur together is n! – [m! . (n – m + 1)!] In how many ways can the letters of word “Culprit” be arranged so that the vowels are never seperate? Solutions : Here n = 7, p = 2 (u, i) Hence by article–5 (above), total ways = p! . (n – p + 1)! = 2! . 6! CIRCULAR PERMUTATIONS Arrangements round a circular table When clockwise and anticlockwise arrangements are considered different then number of circular permutations of n different things taken all at a time is (n–1)! Arrangement of flowers in a garland When clockwise and anticlockwise arrangements are considered same, then required permutations = 1 (n − 1)! 2 Number of circular permutations of n different things taken r at a time = (nPr) / r When clockwise is different than anticlockwise arrangement = (nPr) / (2r) When clockwise & anticlockwise are considered same. In how many ways can 6 ladies and 6 gentlemen be seated at a round table so that no two ladies are next to each other? Solution : Let us arrange the men first. 6 men can be arranged in a circle M in (6 – 1) ! = 5 ! ways. We now have 6 possible places for the six M ladies. The six ladies can be made to occupy six possible places M in 6 ! ways. ∴ Total no. of arrangements is 5 ! × 6 ! = 86400. M M M 20 persons including Mr. A and Mrs. A are invited to a party. In how many ways the host, the hostess and guests can be arranged around a circular table such that the host & the hostess sit together flanked by Mr. A and Mrs. A. Solution : Let Mr. A, host, hostess & Mrs. A form 1 unit, then total 18 + 1 unit = 19 can be arranged round a circular table in 18! ways. More over host & hostess can be seated in two ways & they can be flanked by Mr. & Mrs. A in two ways, hence total ways = 2 × 2 × 18! = 4 × 18! COMBINATION AS SELECTION Combination is the selection of a group which can be made by some or all of number of things without reference to the order. Hence just the selections is combination where as selection cum arrangement is permutation. Generally, number of permutations is larger than number of selections. The number of combinations of n different things taken r at a time is given by nCr or C (n, r) or ⎛ n ⎞ ⎜ r ⎟ . ⎝ ⎠ KEY RESULTS ON nCr (i) n C = n! = r r!(n − r )! n Pr r! ( if r > n, then nCr = 0 ) nCr ∈ N nC0 =n Cn = 1 nC1 = n (v) n Cr =n Cn −r (vi) (vii) n Cr +n Cr −1 =n +1 Cr n.n−1Cr −1 = (n − r + 1). nCr −1 (viii) greatest value of nCr =n Cn / 2 when n is even = nCn +1 or nCn −1 when n is odd (ix) C = n n −1C r r 2 2 r −1 (x) n Cr nCr −1 = n − r + 1 r C0 + C1 + Cn = 2n (xii) C0 + C2 + C4 + .... = C1 + C3 + = 2n−1 (xiii) 2n+1C0 +2n+1 C1 + ........ +2n+1 Cn = 22n (xiv) nCn +n+1 Cn +n+2 Cn + ........ +2n−1 Cn =2n Cn+1 A person wants to host as many different parties as possible by inviting a fixed number of guests from among 21 persons. How many should he invite to each party? Solution : Here n = 21, Let he invites r guests at a time So nCr = 21Cr should have maximum value. Here n is odd, so r = n − 1 or 2 n + 1 2 i.e. = 10 or 11 21C10 = 21C11 is the greatest value of 21Cr SIMPLE APPLICATIONS OF COMBINATION The number of combinations of n different things taken r at a time When k particular things always occur is n–kCr–k . When k particular things never occur is n–kCr . In how many ways can we choose 6 players among 20 contestants when 7 contestants are unfit? Solution : Here n = 20 k = 7 and r = 6 Hence 20–7C6 = 13C6 is the required number of ways Number of combination of n different things selecting atleast one of them is nC1 + nC2 + nCn = 2n – 1 Number of combination of n identical things, taking r (r ≤ n) at a time is 1. Number of ways of selecting none, one or more things out of n alike things is (n + 1) (i.e. 0, 1, 2,................ or n things) Out of (p + q + r + s) things. p are alike of one kind, q are alike of second kind, r are alike of third kind and s are different each, then total number of combination is (p + 1) (q + 1) (r + 1) (2)s – 1 Divisors of N Let N = P α1 .P α2 .P α3 P αk where P1, P2 Pk are different primes and α1, α2 αk are natural numbers; then Total number of divisors of N including 1 & N is (α1 + 1) (α2 + 1) (αk + 1) Sum of these divisors excluding 1 & N is = ( P 0 + P 1 + ........P α1 ) ( P 0 + P 1 + ........P α2 )........... ( P 0 + P 1 + P αk ) 1 1 1 2 2 2 k k k The number of ways in which N can be resolved into two factors which are coprime to each other is 2n–1 where n is the number of different factors in N. Find the number of factors (excluding 1 & itself) of the number 38808. Find the sum of all these factors. In how many ways 38808 can be split in two factors which are co prime? Solution : 38808 = 23.32.72.11. Required number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) – 2 = 70 Sum of these divisors = (20 + 21 + 22 + 23) (30 + 31 + 32) (70 + 71 + 72) (110 + 111) = 15 × 13 × 57 × 12 = 133380 Here n = 4 (because 2, 3, 7 & 11 are coprime) ∴ number of ways in which 38808 can be resolved into two coprime factors is = 24–1 = 23 = 8 Division into groups The number of ways in which (m + n) different things can be divided into two groups containing m and n things (m ≠ n) respectively are m +n Cm .n Cn = (m + n)! m!n! If m = n and group order is not important then 2n! (2)!(n!)2 If m = n and groups are distinct then Generalisation 2n! (n!)2 The number of way in which mn different things can be divided equally among m groups of n each. If order of group is not important (mn)! m!(n!)m If order of group is important = (mn)! (n!)m In how many ways the 52 cards of a pack can be distributed Equally among 4 heaps Equally among 4 players In 4 sets, three of 17 cards each & fourth having 1 card only? Solution : (a) (52)! 4!(13!)4 as groups are identical (b) (52)! (13!)4 as groups are non-identical (c) Ways of selecting 1 card out of of 52 is 52C1 = 52 The remaining 51 cards are to be divided into 3 identical groups = (51)! 3!(17!)3 ∴ Required no is 51! 3!(17!)3 × 52 Arrangement into groups To arrange n different things into r different groups. If blank groups are allowed, then n–r + 1Pn If blank groups are not allowed, then n!.n–1Cr–1 To distribute n different things into r different groups rn – rC1(r–1)n + rC2(r–2)n (–1)r–1 rC–r. OR Coefficient of xn in n!.(ex–1)r (Blank groups are not allowed here) Note : Coefficient of xr in epx = pr r! To distribute n identical things into r different groups If blank groups are allowed, then n+r–1Cr–1 If blank groups are not allowed, then n–1Cr–1 To distribute n identical things in r groups so that no group contains less than l and more than m things (m > l) is coefficient of xn in the expansion of (xl + xl+1...... + xm)r To select r things from a group of n things having p things identical is r ∑ n −p Cr r =0 if r ≤ p and ∑ n− p Cr r =r −p if r > p In how many ways can 5 flags, one for each country can be arranged on top of 3 different buildings so that no building remains empty? Solution : From 8(a) above, the required number of ways are 5!.4C2 = 120 × 6 = 720 In how many ways can 5 flags, one for each country can be distributed over 3 different buildings, so that no building remains empty? Solution : As per article 8(b) above, required ways = 35–3C125 + 3C215 = 243 – 96 + 3 = 246 – 96 = 150 Derangements If n things are arranged in a row, the number of ways in which they can be deranged so that no one of them occupies its original place or no object goes to its scheduled place, is Remark : n!⎜1− 1 + ⎝ 1! 1 − 1 2! 3! + ... + (−1)n 1 ⎞ ⎟ n! ⎠ If r things go to wrong place out of n things, then (n – r) things go to original place (Here r < n) Dn = No. of ways, if all n things go to wrong place and Dr = No of ways, if r things goes to wrong place. If r goes to wrong place out of n, then (n – r) goes to correct places. Then Dn = nCn–rDr. D = ⎛ 1 1 1 r 1 ⎞ Where r r!⎜1− + ⎝ 1! − 2! 3! + ... + (−1) ⎟ r! ⎠ Five letters are written to five persons and five envelops are printed with their addresses. In how many ways the letters can be kept in the envelopes (a) so that none goes to correct envelops (b) atleast three letters go to wrong envelops? Solutions : (a) D = ⎛ 1 1 1 1 1 ⎞ 5 5!⎜1− + − + ⎝ 1! 2! 3! − ⎟ 4! 5! ⎠ = 120 – 120 + 60 – 20 + 5 – 1 = 44 (b) The number of ways in which 3 letters go to wrong envelops = 5C5–3.D3 The number of ways in which 4 letters go to wrong envelops = 5C5–4.D4 The number of way in which all 5 letters go to wrong envelops = 5C5–5.D5 = D5 So required number of ways = 5C2 × D3 + 5C1 × D4 + D5 = 10.3!⎜1− 1 + ⎝ 1! 1 1 ⎞ ⎟ + 2! 3! ⎠ 5.4!⎜1− 1 + ⎝ 1! 1 − 1 + 2! 3! 1 ⎞ ⎟ 4! ⎠ + 44 = 10(3–1) + 5 [12 – 4 + 1] + 44 = 20 + 45 + 44 = 109 Example 1 : Out of 7 consonants and 4 vowels, how many words can be made, each containing 3 consonants and 2 vowels? Solution : The no. of ways of choosing 3 consonants out of 7 consonants is 7C3 while the no. of ways of choosing vowels out of 4 vowels is 4C2. Since each way of selection in the first case may be associated with any of the selections in the second case to form a word, the number of combined groups, each containing consonants and 2 vowels is 7C3 × 4C2. Each of these groups of 5 letters may be arranged among themselves in 5 ! ways. Hence, required no. of words is 7C3 × 4C2 × 5! = 25200. Example 2 : A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups; each group containing 6 questions. He is not permitted to answer more than 5 questions from either group. In how many different ways can he choose the seven questions? Solution : Let the two groups be denoted as groups I and II. The following choices are possible (as re : number of questions) Group I Group II (a) 2 5 (b) 3 4 (c) 4 3 (d) 5 2 The first selection pertains to choice of 2 questions from group I and 5 questions from group II which can be done in 6C2 ⋅ 6C5 = 15 × 6 = 90 ways (by fundamental principle) Similarly 2nd, 3rd and 4th selections can be done in i.e. 300, 300 and 90 ways respectively. ∴ Total no. of ways = 780. 6C3 × 6C4, 6C4 × 6C3 and 6C5 × 6C2 ways respectively, Example 3 : Given 5 different green dyes, four different blue dyes and 3 different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye ? Solution : At least one green dye can be selected out of 5 different green dyes in 25 – 1 ways. 5C1 + 5C2 + + 5C5 ways, i.e. After selecting one or more green dyes, we can select at least one blue dye out of 4 different blue dyes in 24 – 1 ways. Subsequently at least one red dye or no red dye can be chosen in 3C0 + 3C1 + 3C2 + ....+ 3C3 ways = 23 ways ∴ Total no. of combinations of dyes = (25 − 1) ⋅ (24 − 1) ⋅ 23 = 31× 15 × 8 = 3720. Example 4 : How many different numbers can be formed by using all the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places ? Solution : 1 2 3 4 5 6 7 There are 7 digits ; we require 7 digit numbers, which is same as filling 7 places in a row with the 7 digits 1, 2, 3, 4, 3, 2, 1. The odd places are place numbers 1, 3, 5, 7. Odd digits available are 1, 3, 3, 1 ; there are two 1’s and two 3’s. P4 = 4 ! = The 4 odd digits can be arranged at the 4 odd places in 2 ! 2 ! 6 ways 4 Now, there are 3 even places, i.e., place numbers 2, 4, 6 and three digits available, viz., 2, 4, 2. These 3 digits can be arranged at the 3 even places in 3 P3 = 2 ! 3 ways. By fundamental theorem, the required no. of ways is 6 × 3 = 18 ways. Example 5 : Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all the 5 balls. In how many different ways can we place the balls so that no box remains empty? Solution : Each box must contain at least one ball since no box remains empty. Two cases arise : Case I. No. of balls This can be done in 5 C1 × 4C1 × 3C3 = 20 ways But the box which contains 3 balls can be chosen in 3C1 = 3 ways. ∴ Required no. of ways = 20 × 3 = 60 Case II. No. of balls This can be done in 5 C1 × 4C2 × 2C2 = 5 × 6 × 1 = 30 ways But the two boxes containing two balls each can be chosen in 3C2 ways. Box I Box II Box III 1 1 3 Box I Box II Box III 1 2 2 ∴ Required number = 30 × 3 = 90 Since case I and case II are non overlapping the total number of ways = 60 + 90 = 150. Example 6 : Find the sum of all numbers greater than 104 formed by using the digits 0, 2, 4, 6, 8, no digit being repeated in any number. Solution : Required sum = (Sum of all (5 digit) numbers that can be formed by using the given digits 0, 2, 4, 6, 8) – (Sum of the numbers in which ‘0’ occurs in ten thousand’s place) To find the sum of five digit numbers formed by 0, 2, 4, 6, 8 : No. of numbers that can be formed by the given digits = 5 ! = 120 No. of numbers in which a particular digit, say 2, is in a given place (say, unit’s place) If 2 is in unit’s place its value is 2. If 2 is in ten’s place its value is 20. … … … … … … … … If 2 is in ten thousand’s place its value is 20000. ∴ Sum of all five digit numbers = 24 (0 + 2 + 4 + 6 + 8) (10000 + 1000 + 100 + 10 + 1) = 24 × 20 × 11111 Similarly, sum of numbers when ‘0’ is in the ten thousand’s place = 4 ! (2 + 4 + 6 + 8) (1000 + 100 + 10 + 1) = 6 × 20 ×1111 4 ∴ The required sum = 24 × 20 × 11111 – 6 × 20 × 1111 = 20 (266664 – 6666) = 5199960. = 5 ! = 24 5 Example 7 : There are p intermediate railway stations on a railway line from one terminal to another. In how many ways can a train stop at three of these intermediate stations if no two of these stations (where it stops) are to be consecutive? Solution : Since the train does not stop at (p – 3) stations, the problem reduces to the following : In how many ways can three objects be placed among (p – 3) objects in a row such that no two of them are next to each other (at most 1 object is to be placed between any two of these (p – 3) objects). Since there are (p – 2) positions to place the three objects, the required number of ways = p–2C3. Example 8 : A condolence meeting is being held in a hall which has 7 doors, by which mourners enter the hall. One can use any of the 7 doors to enter and can come at any time during the meeting. At each door, a register is kept in which a mourner has to affix his signature while entering the hall. If 200 people attend the meeting, how many different sequences of 7 lists of signatures can arise ? Solution : There are 7 lists, say 1, 2, .......7. Suppose that list i has xi names ; then, x1 + ........ + x7 = 200 where xi ≥ 0 is an integer We need to first find the no. of solutions of this equation. (Note that this does not complete the solution to the question as, list 1 may contain 7 names which no. would remain the same in 7! arrangements of the names) The number of solutions is 200 + 7–1C7–1 = 206C6 But corresponding to any one solution, (x1 ........ x7) (i.e. list j contains xj names) we can have 200 ! arrangements consistent with distribution of xj names to jth list ∴ No. of different sequences of 7 lists = 206C6 × 200 ! = 206 ! 6 ! . ❑ ❑ ❑