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Miscellaneous

Question-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 ?

Solution :-
  In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R.
  Number of ways of selecting 2 vowels out of 3 vowels = ³C₂ = 3!/2!1! = 3
  Number of ways of selecting 3 consonants out of 5 consonants = ⁵C₃ = 5!/3!2! = 10
  Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30
  Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.
  Hence, required number of different words = 30 × 5! = 3600
   

Question-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?

Solution :-
  In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.
  Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. 
  Then, the permutations of these 2 objects taken all at a time are counted. 
  This number would be ²P₂ Corresponding to each of these permutations, 
  there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time. 
  Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440
   

Question-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 ?

Solution :-
  A committee of 7 has to be formed from 9 boys and 4 girls.
(i) Since exactly 3 girls are to be there in every committee, each committee must consist of (7 – 3) = 4 boys only.
  Thus, in this case, required number of ways = ⁴C₃ x ⁹C₄ = 4!/3!1! x 9!/4!5! = 504
  
(ii) Since at least 3 girls are to be there in every committee, the committee can consist of
  3 girls and 4 boys can be selected in ⁴C₃ x ⁹C₄ ways.
  4 girls and 3 boys can be selected in ⁴C₄ x ⁹C₃ ways .
  Therefore, in this case, required number of ways = ⁴C₃ x ⁹C₄ + ⁴C₄ x ⁹C₃ = 504 + 84 = 588
    
(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of
  3 girls and 4 boys can be selected in ⁴C₃ x ⁹C₄ ways.
  2 girls and 5 boys can be selected in ⁴C₂ x ⁹C₅ ways.
  1 girl and 6 boys can be selected in ⁴C₁ x ⁹C₆ ways.
  No girl and 7 boys can be selected in ⁴C₀ x ⁹C₇ ways. 
  Therefore, in this case, required number of ways 
= ⁴C₃ x ⁹C₄ + ⁴C₂ x ⁹C₅ + ⁴C₁ x ⁹C₆ + ⁴C₀ x ⁹C₇ 
= 504 + 756 + 336 + 36
= 1632
   

Question-4 :-  If the different permutations of all the letter 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 ?

Solution :-
  In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times and 
  all the other letters appear only once.
  The words that will be listed before the words starting with E in a dictionary will be the words that start with A only.
  Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, 
  and then the remaining 10 letters taken all at a time are rearranged.
  Since there are 2 Is and 2 Ns in the remaining 10 letters,
  Number of words starting with A = 10!/2!2! = 907200
  Thus, the required numbers of words is 907200.
   

Question-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 ?

Solution :-
  A number is divisible by 10 if its units digits is 0.
  Therefore, 0 is fixed at the units place.
  Therefore, there will be as many ways as there are ways of filling 5 vacant places
  in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).
  The 5 vacant places can be filled in 5! ways.
  Hence, required number of 6-digit numbers = 5! = 120 
   

Question-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 ?

Solution :-
  2 different vowels and 2 different consonants are to be selected from the English alphabet.
  Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet = ⁵C₂ = 5!/2!3! = 10
  Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet = ²¹C₂ = 21!/2!19! = 210
  Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100
  Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
  Therefore, required number of words = 2100 × 4! = 50400
   

Question-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 ?

Solution :-
  It is given that the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively.
  A student has to attempt 8 questions, selecting at least 3 from each part. This can be done as follows.
  3 questions from part I and 5 questions from part II can be selected in ⁵C₃ x ⁷C₅ ways.
  4 questions from part I and 4 questions from part II can be selected in ⁵C₄ x ⁷C₄ ways.
  5 questions from part I and 3 questions from part II can be selected in ⁵C₅ x ⁷C₃ ways. 
  Thus, required number of ways of selecting questions 
= ⁵C₃ x ⁷C₅ + ⁵C₄ x ⁷C₄ + ⁵C₅ x ⁷C₃
= 210 + 175 + 35 
= 420
   

Question-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.

Solution :-
  From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.
  In a deck of 52 cards, there are 4 kings.
  1 king can be selected out of 4 kings in ⁴C₁ ways.
  4 cards out of the remaining 48 cards can be selected in ⁴⁸C₄ ways.
  Thus, the required number of 5-card combinations is ⁴C₁ x ⁴⁸C₄ .
   

Question-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 ?

Solution :-
  5 men and 4 women are to be seated in a row such that the women occupy the even places.
  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 x M x M x M x M
  Therefore, the women can be seated in 4! ways.
  Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 2880.
   

Question-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 ?

Solution :-
  From the class of 25 students, 10 are to be chosen for an excursion party.
  Since there are 3 students who decide that either all of them will join or none of them will join, there are two cases. 
  Case I: All the three students join.
  Then, the remaining 7 students can be chosen from the remaining 22 students in ²²C₇ ways.
  Case II: None of the three students join.
  Then, 10 students can be chosen from the remaining 22 students in ²²C₁₀ ways.
  Thus, required number of ways of choosing the excursion party is ²²C₇ x ²²C₁₀.
   

Question-11 :-  In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ?

Solution :-
  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.
  Since all the words have to be arranged in such a way that all the Ss are together, 
  SSSS is treated as a single object for the time being. 
  This single object together with the remaining 9 objects will account for 10 objects.
  These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in 10!/3!2!2! ways.
  Thus, required number of ways of arranging the letters of the given word 10!/3!2!2! = 151200
   
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