EXERCISE - 02 BRAIN TEASERS UNIT # 12 PERMUTATION & COMBINATION PERMUTATION & COMBINATION AND PROBABILITY EXERCISE - 01 CHECK YOUR GRASP
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1 EXERCISE - 01 CHECK YOUR GRASP 1. Total possible words words do not begi n or terminate with vowel Total words! 10 Words which do not begin and terminate with vowel 1 6 Desired words : II-Method words which begin with vowel (A/I 4! 48 say n(a Similarly words terminating with vowel 4! 48 say n(b Now exclude words which begin as well as terminates with vowel n(a B Desired number of words : ( n(a B n(a + n(b n(a B. Number of selections of 4 consonants out of is C 4 Number of selections of vowels from 4 is 4 Arrangement of words in 6! Desired words : C 4 4 6! Make cases when all boxes are filled by: Case 1 : Case : Case : identical red balls C 1 way 4 identical red balls and 1 blue ball blue and red balls i.e. xrxrx 4 gaps, for blue balls 4 6 Case 4 : red and blue balls i.e. xrxrx gaps, blue balls C 1 way Total number of are EEQUU Words starting with E 4!! Words starting with QE!! next word will be QUEEU 1 and finally QUEUE 1 Rank is th 1. For a number to be divisible by, or 0 should be at units place. Unit place can be filled by 6! Remaining digits can be filled in!!. Total 6!!! But these arrangements also include cases where 0 is at millions place and at units place, which are undesirable cases! (undesirable!! subtract it from total. Desired 6!!!!!! 110 EXERCISE - 0 BRAIN TEASERS 1. I H Indian husband, I W Indian wife A H American husband, A W American wife Case 1 : Hand shaking occuring between same nationals and same genders (M/F I H I H 10 Similarly for I W I W, A W A W, A H A H Total Case : All other possible hand shakes I H I w A H A w 40 Hence total number of handshakes ( (0 1 UNIT # 1 PERMUTATION & COMBINATION AND PROBABILITY PERMUTATION & COMBINATION Me th od -II Total number of handshakes possible 0 undesirable handshakes : Hence, Desired 0 ( Case 1 : When all n red balls are taken but no green ball. Only 1 arrangement is possible. Case : n red balls and 1 green balls n 1 Number of arrangement n 1 Case : n red balls and green balls n Number of arrangement n case m + 1 : n red balls and m green balls n m Number of arrangements n m add all cases
2 1 + n 1 n 1 + n n n m n m n+1 C 0 + n+1 + n n+m C m Similarly if we start from A towards B we get another 4 paths. A n+ + n n+m C m ( n C r + n C r 1 n+1 C r n+ + n+ C n+m C m A D C B A Finally we get the sum as : m+n+1 C m. Maximum number of matches possible, if India win the series are 9 Out of these 9 matches we are required to choose matches that India may win : Total : 9 C 6. Paths are shown as :- A A C D B A B A D A 4 paths B A C A Similarly if we start from A towards B Again 4 paths Total different paths 4 1 II Method 1. Case 1 : Mr. B & Miss C are in committee and Mr. A is excluded (Men (Women Case : Mr. B not there : C 100 (Men Total (Women EXERCISE - 0 MISCELLANEOUS TYPE QUESTIONS Match the column : 1. (A Each box (say B 1, B, B will have at least one ball. (B (C (D Now the for placing other identical balls in different boxes are :- ( 1! 6! ( 1! (n r 1! n! (r 1! Case 1 : balls can be divided in groups having balls each in boxes and 1 ball for in third box (,, 1 :! (1!(!! 1 Case : Division can also be in one box and 1 each in remaining boxes (, 1, 1! :! (1!! 10 Hence total Only arrangements are possible. (1 balls each in boxes & remaining ball in other box (,, 1 ( balls in 1 box and 1 ball each in other boxes (, 1, 1 Same cases as that of in part B with arrangements. Comprehension # 0 1. Exponent of in 100! exponent of in 0! 0 8 Exponent of in 100 C 0 100! 0!0! exponent of will be 0.. Product of 's & 's constitute 0's at the end of a number No. of 0's in 108! exponent of in 108! (Note that exponent of will be more than exponent of in 108! As 1., here we have to calculate exponent of and exponent of in 100! exponent of exponent of 48 Now, 1 we require two 's & one exponent of will give us the exponent of 1 in 100! i.e. 48
3 EXERCISE - 04[A]. Selecting horses out of ABC A'B'C' is 6 C When AA' is al selected among (ABC A'B'C' Remaining (BB'CC' can be selected in 4 similarly, when BB' and CC' is selected Undesirable will be ( 4 using, total undesirable desired we get ( 6 C ( 4 This is selection of horses among (ABC A'B'C' under given condition. Remaining can be selected in 10 C. Hence, desired will be [ 6 C 4 ] 10 C 9 Method II : Select one horse each from AA', BB' and CC' hence. Now select horses from remaining 10 horses in 10 C. Total 10 C. Case 1 : When all digits are same : 9 (excluding 0 Case : When digits are different and 0 excluded. (a (b (c Selecting numbers in 9 Each digit can be filled in hence 4 way Undesirable case : when a particular digit is same ( 1 (case 1 9 ( 4 Case : When digits are different and 0 is included (a (b (c other digit can be chosen in 9 0 can't be placed at ten thousand's place, hence selected digit should be fixed at this place remaining digits can be filled with. Undesirable case : When all the 4 digits gets filled with selected digit only (0 not included 1 way hence no. of will be : 9 ( 1 Total desired : ( ( 1 6. Husband H, Wife W Given : Relatives of husband (H (a Ladies (L H 4 (b Gentlemen (G H Relatives of Wife (W (a Ladies (L W (b Gentlemen (G W 4 Case 1 : Selecting (L H and (G W : 4 C 4 C 16 CONCEPTUAL SUBJECTIVE EXERCISE Case : Selecting (G H and (L W : C C 1 Case : Selecting (L H & 1G H & (1L w & G w : Case 4 : Selecting (1L H & G H & (L W & 1G w : Add all cases we get : Step 1 st : Arrange boys in! Step nd : Select gaps from 6 gaps for 4 girls (girls for each gap in 6. Step rd : Select girls to sit in one of the gaps and other in remaining selected gaps 4 Step 4 : Arrange 1 st, girls in! and other in! Hence, total! Distribute 1 candies among. Ram (R + Shyam(S + Ghanshyam(G + Balram(B with condition given : R+S+G+B1&R & S After giving to Shyam, remaining candies 1 1 Now distribute 1 candies in R, S, G, B in C In 16 C, we have to remove undesirable, when R > Undesirable : R > R 6 give at least 6 to R and to S and distribute remaining between R, S, G, B 1 ( + 6 remaining can be distributed between R, S, G, B in 10 C are the undesirable cases Desired 16 C 10 C C 1 8. Total number of Put exactly one ball in its box and then dearrange remaining balls. L NM 4! !!! 4! 9 4 O QP
4 EXERCISE - 04[B] BRAIN STORMING SUBJECTIVE EXERCISE. (i Total 10! undesirable cases : when Americans are together (A 1 A or two British are together (B 1 B or two Chinese are together ( we plot them on Venn diagram : we use, AA 1 BB 1 C n(a 1 A B 1 B n (A 1 A + n(b 1 B + n( n[(a 1 A (B 1 B ] n [(B 1 B ( ] n [( (A 1 A ] + n [(A 1 A (B 1 B ( ] where n(a 1 A denotes when Americans are together 9!! 6. (a Selection of r things out of n +1 different things Selection of r things out of n +1 different things, when a particular thing is excluded + a particular thing is included. (b Selection of r things out of not m + n different things can be made by selecting x thing from m and y thing from such that x +y r & (x, y (0, r, (1, r 1, (, r,...(r, 0 9. clerks who prefer Bombay are to be sent to different companies in Bombay, and Out of remaining clerks (excluding clerks who prefer for outside clerks are chosen in. Now these 4 can be sent to different companies into groups of each in 4 4 Now for outside companies we have 6 clerks remaining we select them as ( for each company (ii correspondingly for B 1 B & n[(a 1 A (B 1 B ] denotes when Americans and Britishmen are together 8!!! correspondingly same for others. n[(a 1 A (B 1 B ( ] denotes when Americans, Britishmen and Chinese are together!!!! 86 Put values we get n(a 1 A B 1 B 9!! 8! + 8! 8!(4 These are undesired Desired 10! 8! (4 8!(4 Now they are on a round table Total (n 1! (10 1! 9! Undesired : n(a 1 A B 1 B 8!!!!! + 6!!!! 6! 4 [ + ] 6! 60 Desired 9! 6! 60 (44 6! 6 4 Desired ( 4 ( Total cases of selecting 8 men out of 11 is 11 C 8 Undesirable case : When all the men, who can only row on stroke side are selected then other men can be selected in 6 C. Hence, 11 C 8 6 C 14 Ans. 1. Step 1 st : Select lines out of n lines in n to get a point (say p. Step- nd : Now select another lines in n, to get another point (say Q Step- rd : When P and Q are joined we get a fresh line. P Q Fresh line But when we select P first then Q and Q first then P we get same line. C C n n Fresh lines
5 EXERCISE - 0 [A] 1. Numbers greater than 1000 and less than or equal to 4000 will be of 4 digits and will have either 1 (except 1000 or or in the first place with 0 in each of the remaining places. After fixing 1 st place, the second place can be filled by any of the numbers. Similarly third place can be filled up in and 4 th place can be filled up in. Thus there will be 1 in which 1 will be in first place but this include 1000 also hence there will be 14 numbers having 1 in the first place. Similarly 1 for each or. One number will be in which 4 in the first place and i.e Hence the required numbers are We know that a five digit number is divisible by, if and only if sum of its digits ( 1 is divisible by. Therefore we should not use 0 or while forming the five digit numbers. Now, (i in case we do not use 0 the five digit number can be formed (from the digit 1,,, 4, in P. (ii In case we do not use, the five digit number can be formed (from the digit 0, 1,, 4, in P 4 P 4! 4! The total number of such digit number P + ( P 4 P No. of in which 6 men can be arranged at a round table (6 1! Now women can be arranged in 6!. Total Number of 6!!. As for given question two cases are possible (i Selecting 4 out of first question and 6 out of remaining 8 questions C 4 8 C choices (ii Selecting out of first questions and out of remaining 8 questions C 8 C 6 choices. Total no. of choices Number of to arrange in which vowels are in alphabetical order 6!! Number of n 1 C r C 1 1 JEE-[MAIN] : PREVIOUS YEAR QUESTIONS C + 10 C 4 8 1! 1. Number of (4!.! 1!! (4! 1 4. The no. of to select 4 novels & 1 dictionary from 6 different novels & different dictionary are 6 C 4 and to arrange these things in shelf so that dictionary is al in middle D are 4! Required No. of 6 C 4 4! Urn A Red balls Urn B 9 Blue balls So the number of selection of balls from urn A & B each B 1 + B + B + B 4 10 St - 1 : B 1 1, B 1, B 1, B 4 1 so no. of negative integers solution of equation x 1 + x + x + x C C St - : selction of places from out of 9 places 9 C Both statements are true and correct explaination 1. N 10 C 6 C N W 10, G 9, B selection of one or more balls ( (9 + 1 ( (A, B C + 8 C C C T n n C n+1 C n C 10 (n + 1 n (n 1 n(n 1 (n 60 n(n 1 0 n
6 EXERCISE - 0 [B] 4. x + y < 1 x + y 0 x + y 18 ( x > 0 & y > 0 Introducing new variable t x + y + t 18 x>0 y>0 Now dividing 18 identical things among persons C JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 Balls can be distributed as 1, 1, or 1,, to each person. When 1, 1, balls are distributed to each person, then total number of :! 1..! 60 1! 1!!! When 1,, balls are distributed to each person, then total number of :! 1..! 90 1!!!! total Paragraph for Question 1 and 14 : For a n The first digit should be 1. Total number of of distributing n objects into n groups, each containing n objects ( n! n ( n! n!. n! ( n! ( n! n integer (Since number of are al integer 8. Since, r, s, t are prime numbers. Selection of p and q are as under p q number of r 0 r 1 way r 1 r 1 way r r 0, r 1, r Total number of to select r s 0 s 4 1 way s 1 s 4 1 way s s 4 1 way s s 4 1 way s 4 s 0, s 1, s, s, s 4 Total number of to select s 9. Similarly total number of to select t number of Ans. (D Case- I : The number of elements in the pairs can be 1,1; 1,; 1,,;, C. C C + Case- II : Number of pairs with as one of subsets 4 16 Total pairs For b n 1 _... 1 n Places Last digit is 1. so b n is equal to number of of a n 1 (i.e. remaining (n 1 places For c n b n a n 1 Last digit is 0 so second last digit must be 1 So c n a n b n + c n a n So a n a n 1 + a n Similarly b n b n 1 + b n 1. A n s. ( B a 1 1, a So a, a 4 a 8 b 6 a A ns.(a a n a n 1 + a n put n 1 a 1 a 16 + a 1 c n c n 1 + c n So put n 1 c 1 c 16 + c 1 b n b n 1 + b n put n 1 b 1 b 16 + b 1 a 1 a 16 + a 1 (A is correct (B is incorrect (C is incorrect while (D says a 1 a 1 + a 1 (D is incorrect
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