Permutation and Combination Questions and Answers PDF
Latest Permutation and Combination MCQ Objective QuestionsPermutation and Combination MCQ Question 1:From a total of 25 players, a cricket team has to be formed in which 7 batsmen and 4 ballers are required. If 40% of the total players are ballers then find the number of combinations in which, team can be formed ? Show
Answer (Detailed Solution Below)Option 2 : 8535 Given: Total number of players = 25 Calculation: Total number of ballers = 25 × 40/100 = 10 Then, number of batsman = 25 - 10 = 15 Now, according to question Required number of combination = 15C7 + 10C4 = {(8 × 9 × 10 × 11 × 12 × 13 × 14 × 15)/(1 × 2 × 3 × 4 × 5 × 6 × 7 × 8)} + {(5 × 6 × 7 × 8 × 9 × 10)/(2 × 3 × 4 × 5 × 6)} = 6435 + 2100 = 8535 ∴ The required value is 8535. Permutation and Combination MCQ Question 2:A teacher and headmaster are chosen out of a group having 15 persons. How many ways are there?
Answer (Detailed Solution Below)Option 1 : 210 Given: Total of 15 persons are in the group Concept used: nCr = n!/(n – r)! Where, n = number of items, r = how many items are taken at a time Calculation: Total number of ways = 15C1 × 14C1 ⇒ 210 ∴ Required answer is 210. Permutation and Combination MCQ Question 3:There are 5 boys and 4 girls and they are sitting on a circular table. In how many ways they can sit?
Answer (Detailed Solution Below)Option 4 : 40320 Given: There are 5 boys and 4 girls and they are sitting on a circular table. Formula used: Number of ways in which n peoples can be arranged on a circular table = (n - 1)! Calculation: According to the formula, we have Required number of arrangements = (9 - 1)! = 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 = 40320 ∴ The required number is 40320. Permutation and Combination MCQ Question 4:In how many different ways can the letters of the word “PHONE” be arranged so that the vowels may occupy only odd positions?
Answer (Detailed Solution Below)Option 1 : 36 Concept used: nPr = n!(n – r)! Calculation: Number of ways of arranging the vowels in odd positions = 3P2 = 6 Number of ways of arranging the remaining letters =3! = 6 Required number of ways = 6 × 6 = 36 ∴ Required answer is 36 Permutation and Combination MCQ Question 5:A team of 11 players is to be selected out of 6 defenders, 4 mid-fielders and 5 strikers. Find the number of ways of selecting at most 3 strikers in the team?
Answer (Detailed Solution Below)Option 2 : 450 Given: There are 6 defenders, 4 mid-fielders, and 5 strikers Concept used: nCr = n!/r!(n – r)! Calculation: A team of 11 players consisting of at most 3 strikers can be formed in the following ways. Case 1: When 3 strikers are selected 5C3 × 10C8 ⇒ (5 × 4/2) × (10 × 9/2) ⇒ 10 × 45 = 450 ∴ Required answer is 450 How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 (repetition of digits is allowed)?
Answer (Detailed Solution Below)Option 2 : 9 ⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3. ∴ 9 possible two-digit numbers can be formed. The 9 possible two-digit numbers are: 33, 35, 37, 53, 55, 57, 73, 75, 77 In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?
Answer (Detailed Solution Below)Option 4 : 15120 Given: The given number is 'GEOGRAPHY' Calculation: The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA). Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different. Now, The number of ways to arrange these letters = 7!/2! ⇒ 7 × 6 × 5 × 4 × 3 = 2520 In the 3 vowels(EOA), all vowels are different The number of ways to arrange these vowels = 3! ⇒ 3 × 2 × 1 = 6 Now, The required number of ways = 2520 × 6 ⇒ 15120 ∴ The required number of ways is 15120. How many 3 digit odd numbers can be formed from the digits 5, 6, 7, 8, 9, if the digits can be repeated
Answer (Detailed Solution Below)Option 2 : 75 Given: 5, 6, 7, 8, 9 are the digits to form 3 digit number Calculation: Let us take the 3digit number as H T U (Hundreds, tens, unit digit) respectively To make 3 digit number as odd 5, 7, 9 are only possibly be used in the unit digit place In hundreds and tens place all 5 digits are possible Number of ways for unit digit = 3 Number of ways for tens digit = 5 Number of ways for hundreds digit = 5 Number of 3 digits odd number = 3 × 5 × 5 = 75 ∴ 75 Three-digit odd numbers can be formed from the digits 5, 6, 7, 8, 9 if the digits can be repeated In how many ways can the word CHRISTMAS be arranged so that the letters C and M are never adjacent?
Answer (Detailed Solution Below)Option 1 : 8! × (7/2) Given: Different words from CHRISTMAS have to be formed. Formula: Words in which letters C and M are never adjacent = All cases – words having C and M together. Calculation: ⇒ Total number of words = 9!/2! (Division of 2! As S is repeated) Let us assume C and M to be one unit. Then, letters can be arranged in 8! Ways. C and M can be arranged in 2! Ways. Letter S is repeated, so total number of ways will be divided by 2! ⇒ Number of words with C and M adjacent= 8!/2! × 2! = 8! Words in which letters C and M are never adjacent = 9!/2! – 8! = 8! × (9/2 - 1) = 8! × (7/2) How many four-digit numbers can be formed with digits 2, 5, 6, 7 and 8? (Repeating digits are not allowed)
Answer (Detailed Solution Below)Option 1 : 120 Given: 5 number are given 2, 5, 6, 7 and 8 Four-digit number without repetition Formula used: Permutation for no repetition = \(\frac{n!}{(n \ - \ r)!}\) Where n = total possible numbers r = required number Calculation: Here the total possible number n = 5 And Required number r = 4 Applying the formula \(\frac{5!}{(5\ - \ 4)!}\) ⇒ 5! ⇒ 5 × 4 × 3 × 2 × 1 = 120 ∴ There will 120 possible four-digit number. In how many ways can we sort the letters of the word MANAGEMENT so that the comparative position of vowels and consonants remains the same as in MANAGEMENT.
Answer (Detailed Solution Below)Option 4 : 1080 Given: Word = MANAGEMENT Calculation: Vowel occupy 4 places then !4 ∵ A and E are repeated then !4/(!2 × !2) Consonant occupy 6 places then !6 ⇒ M and N are repeated then !6/(!2 × !2) ∴ \(\frac{{!4}}{{!2\; × \;!2}}\; × \;\frac{{!6}}{{!2\; × \;!2}} = \;\frac{{4\; × 3\; × 2\; × 1}}{{2\; × 1\; × 2\; × 1}}\; × \frac{{6\; × 5\; × 4\; × 3\; × 2\; × 1}}{{2\; × 1\; × 2\; × 1}}\) = 6 × 180 = 1080 In how many different ways can the letters of the word 'FIGHT' be arranged?
Answer (Detailed Solution Below)Option 2 : 120 Given Total alphabets in word 'FIGHT' = 5 Concept Used Total number ways of arrangement = n! Calculation The number of different ways of arrangement of n different words (without repetition) = 5! ⇒ 5 × 4 × 3 × 2 × 1 = 120 ∴ The required answer is 120 How many different 6-digit numbers can be formed from the digits 4, 5, 2, 1, 8, 9 ?
Answer (Detailed Solution Below)Option 3 : 720 The given digits are 4, 5, 2, 1, 8, 9 Therefore required number of ways = \(\frac{n!}{n_1!n_2!...n_k!}\) \(\frac{6!}{(1!)(1!)(1!)(1!)(1!)(1!)}=720\) Hence, the correct answer is 720. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that atleast 3 men are there on the committee. In how many ways can it be done ?
Answer (Detailed Solution Below)Option 4 : 756 Given: (7 men + 6 women) 5 persons are to be chosen for a committee. Formula used: nCr = n!/(n - r)! r! Calculation: Ways in which at least 3 men are selected; ⇒ 3 men + 2 women ⇒ 4 men + 1 woman ⇒ 5 men + 0 woman Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0 ⇒ 7!/(3! × 4!) × 6!/(2! × 4!) + 7!/(4! × 3!) × 6!/(1! × 5!) + 7!/(5! × 2!) × 6!/(6!× 0!) ⇒ 35 × 15 + 35 × 6 + 21 ⇒ 735 + 21 = 756 ∴ The required no of ways = 756.
How many arrangements can be made using the word TESTBOOK so that all the vowels come together
Answer (Detailed Solution Below)Option 2 : 1080
Consider all the vowels together as one unit like TSTBK (EOO) Now, total number of units is (5 + 1) = 6. the units can arrange themselves in 6! ways. the vowels (EOO) can arrange themselves in 3! ways. Now both letter 'T' and 'O' repeat two times So, the total number of ways = 6! × 3!/(2! × 2!) = 4320/4 = 1080 Find the number of ways of the word ANUBHAW can be arranged.
Answer (Detailed Solution Below)Option 4 : 2520 Concept used: Number of all permutations of n things = n! Calculation: The word ANUBHAW contain 7 letters but the letter A appears two times ⇒ No. of way = 7!/2! ⇒ (7 × 6 × 5 × 4 × 3 × 2!)/2! ⇒ 7 × 6 × 5 × 4 × 3 ⇒ 2520 ∴ The word ANUBHAW can be rearranged in 2520 ways In how many ways can the letters of the word LEADER be arranged?
Answer (Detailed Solution Below)Option 3 : 360 Given Word = LEADER Concept A word having 'n' letters in which 'a' letters are repeated can be written in n!/a! ways. Calculation Number of letters in LEADER = 6 Number of repeated letters = 2 ∴ Required ways = 6!/2! = 6 × 5 × 4 × 3 = 360 A team of 3 doctors has to be selected having at least one man. If there are a total of 4 men and 3 women, in how many ways the team can be formed?
Answer (Detailed Solution Below)Option 2 : 34 In the team, there can be one man, two men or three men. Number of ways of selecting one man and two women = 4C1 × 3C2 = 4 × 3 = 12 Number of ways of selecting two men and one woman = 4C2 × 3C1 = 6 × 3 = 18 Number of ways of selecting three men = 4C3 = 4 ∴ Total number of ways = 12 + 18 + 4 = 34 Alternate Method Number of ways of selecting at least one man in the team = total ways of selecting - number of ways of selecting only women ⇒ 7C3 - 3C3 ⇒ 35 - 1 ⇒ 34 In how many different ways the letters of the word ‘MENHIR’ be arranged so that the vowel never comes together?
Answer (Detailed Solution Below)Option 3 : 480 Given: The given word = ‘MENHIR’ Concept used: Assume all the vowels equal to one letter Formula used: If a word has n letters, then the number of different ways to arrange the letter is Case1:- When no repetition of letters takes place = n! Case2:- When r1, r2, r3, … rn repeated letters = n!/(r1! r2! … rn!) Calculation: The given word = ‘MENHIR’ Total number of letters in the given word = 6 The total number of ways to arrange ‘MENHIR’ = 6! = 720 Number of vowel = 2 Total number of letters in the given word after taking two vowels as one = 5 The number of ways to arrange the given word = 5! The vowel can be arranged in 2! Ways Total number of ways to arrange the word when the vowel is together = 5! × 2! = 240 Total number of ways = 720 – 240 = 480 ∴ The number of ways to arrange the given word when a vowel is never together is 480 In how many ways can three men and five boys be seated in a linear arrangement so that all the men sit together?
Answer (Detailed Solution Below)Option 4 : 4320 Given: Three men and five boys are to be seated in a linear fashion so that men sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 The number of ways of arrangements of 10 persons in four chairs is -
Answer (Detailed Solution Below)Option 2 : 5040 Given: The number of ways of arrangements of 10 persons in four chairs Formula used: nPr = n!/(n – r)! Where, n = Number of persons r = Number of chairs Calculation: According to the question nPr = n!/(n – r)! ⇒ 10!/(10 – 4)! ⇒ 10!/6! ⇒ (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(6 × 5 × 4 × 3 × 2 × 1) ⇒ (10 × 9 × 8 × 7) ⇒ 5040 ∴ The required value is 5040 A box contains 5 red, 4 white and 3 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours.
Answer (Detailed Solution Below)Option 3 : 60 1 red ball can be selected in 5C1 ways 1 white ball can be selected in 4C1 ways 1 blue ball can be selected in 3C1 ways ∴ Total number of ways = 5C1 × 4C1 × 3C1 = 5 × 4 × 3 = 60 A box contains 4 red balls, 7 green balls and 5 yellow balls. In how many ways 4 balls can be drawn from the box?
Answer (Detailed Solution Below)Option 4 : 1820 Given: Total number of balls = (4 + 7 + 5) = 16 Concept used: When any specified colour of the ball is not mentioned at the time of drawing the ball then we can pick any ball from the box. Formula used: The different combinations of ‘n’ distinct objects taken ‘r' at a time are, C(n, r) = n!/{r! (n – r)!} Calculation: Total number of balls, (n) = 16 Number of balls to be drawn, (r) = 4 Number of ways to draw balls = C(16, 4) ⇒ 16!/(12! 4!) ⇒ (16 × 15 × 14 × 13)/(4 × 3 × 2) ⇒ 1820 ∴ The number of ways to draw balls from the box is 1820 From a group of 8 men and 7 women, in how many ways can 6 men and 4 women be selected?
Answer (Detailed Solution Below)Option 4 : 980 Given: Group of 8 men and 7 women. Formula used: \({{\rm{n}}_{{{\rm{C}}_{r\;}} = \;\frac{{{\rm{n}}!}}{{{\rm{r}}!{\rm{\;}}\left( {{\rm{n\;}} - {\rm{\;r}}} \right)!{\rm{\;}}}}}}\) Calculation: 8C6× 7C4 out of total 8 men, 6 men selected = 8C6 out of total 7 women, 4 women selected = 7C4 8C6 × 7C4 \(\Rightarrow \;\frac{{8!}}{{6!\; × \;2!}}\; × \;\frac{{7!}}{{4!\; × \;3!}}\) \(\Rightarrow \;\frac{{8\; × \;7\; × \;6!}}{{6!\; × \;2}}\; × \;\frac{{7\; × \;6\; × \;5\; × \;4!}}{{4!\; × \;3\; × 2}}\) ⇒ 28 × 35 ⇒ 980 ∴ 6 men and 4 women are selected in 980 ways. In how many ways can a group of 2 men and 3 women be selected out of a total of 8 men and 10 women?
Answer (Detailed Solution Below)Option 3 : 3360 Given: Number of women = 10 Number of women to be selected = 3 Number of men = 8 Number of men to be selected = 2 Concept used: The number of ways in which 'r' men can be selected from the group of n men = nCr = n!/[r! × (n-r)!] Calculation: Number of ways in which 2 men can be selected from 8 men = 8C2 = 8!/[2! × (8 - 2)!] = 8!/(2! × 6!) = 28 Number of ways in which 3 women can be selected from 10 women 10C3 = 10!/[3! × (10 - 3)!] = 10!/(3! × 7!) = 120 Number of ways in which 2 men and 3 women can be chosen = 8C2 × 10C3 = 28 × 120 = 3360 ∴ The number of ways in which 2 men and 3 women can be selected is 3360. How do you solve permutation and combination questions?What is the formula for permutations and combinations? The formula for permutations is: nPr = n!/(n-r)! The formula for combinations is: nCr = n!/[r! (n-r)!]
What is the difference between permutation and combination PDF?Permutation refers to the different ways of arranging a set of objects in a sequential order. Combination refers to several ways of choosing items from a large set of objects, such that their order does not matters.
What is permutation and combination problems?Permutations are for lists (order matters) and combinations are for groups (order doesn't matter). You know, a "combination lock" should really be called a "permutation lock". The order you put the numbers in matters. A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.
How do you solve combination problems?To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial.
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