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Math Statistic study about permutation and combination.

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- 1. Math StatisticCambridge University PressBy: danica pPermutationandCombination
- 2. Permutation, what s it?Permutation is an arrangement of elements of a set in which the order of theelements considered.Means, it explain the number of ways in which a subset of objects can beselected from a given set of objects, where order is important.Example:Given the set of three letters, {A, B, C}, how many possibilities are there for selectingany two letters where order is important?Answer: (AB, AC, BC, BA, CA, CB) there is 6 arrangements are possible.from A:A-B, A-C (2 arrangements)From B:B-A, B-C (2 arrangements)From C:C-A, C-B (2 arrangements) Total: 2*3 = 6 arrangementsNotice!PERMUTATION (P)=ARRANGEMENT !
- 3. Factorial Formula of Permutation!.( )!n rnPn r Notation of n = n!n! = n*(n-1)*(n-2)*.......*2*1e.g. 9!=9*8*7*6*5*4*3*2*1* = multiplyExample:15 runners are hoping to take part in a marathon competition, but the track has only 4lines. How many ways can 4 of the 15 runers be assigned to lanes?Answer: Using Factorial formula of P,15P4 = = 327601*2*3*41*2*3*4*5*6*7*8*9*10*11*12*13*14*15)!1115(!15You can using your scientificcalcuator, it simple, you enter number n,then shift-x-then enter the number of r, try it!15-shift-x-P- then, 4, so it became to 15P4 =32760
- 4. Then, what about a combination?Combination is an arrangement of elements of a set in which the order ofelements is not considered.Means: The number of ways in which a subset of objects can be selected from agiven set of objects, where order is not important.Example:Given the set of three letters, {A, B, C}, how many possibilities are there forselecting any two letters where order is not important?Answer: (AB, AC, BC) 3 combinations.each alphabet has different couple combination, and that is:A can together with B,A can together with C,B can together with C, total: 3Notice!COMBINATION (C)=CHOOSE!
- 5. Factorial Formula for Combinations!.! !( )!n rn rP nCr r n rExample:The manager of a football team has a squad of 18 players. He needs to choose 11 to playin a match. How many possible teams can be choosen?Answer: Using Factional formula of C,18C11= = 31824!7!*11!18)!1118(!11!18!111118PYou can using your scientificcalcuator, it simple, you enter number n, thenshift-( )-then enter the number of r, try it!18-shift-( )-P- then, 11 so it became to18C11= 31824
- 6. Difference between Permutation and Combination
- 7. Exercises:1. From a group of 7 men and 6 women, five persons are to be selected toform a committee so that at least 3 men are there on the committee. In howmany ways can it be done?A.564 B.645 C.735 D.756 E.None of theseAnswer: Option DExplanation:We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)=7 x 6 x 5x6 x 5+(7C3 x 6C1) + (7C2)3 x 2 x 12 x 1= 525 +7 x 6 x 5x 6+7 x 63 x 2 x 12 x 1= (525 + 210+ 21)= 756.2. A box contains 2 white balls, 3 black balls and 4 red balls. In how manyways can 3 balls be drawn from the box, if at least one black ball is to beincluded in the draw?A.32 B.48 C.64 D.96 E.None of theseAnswer: Option CExplanation:We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3black).Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)=3 x6 x 5+3 x 2x 6+12 x 12 x 1= (45 + 18 + 1)= 64.
- 8. 3. In how many different ways can the letters of the word LEADING bearranged in such a way that the vowels always come together?A.360 B.480 C.720 D.5040 E.None of theseAnswer: Option CExplanation:The word LEADING has 7 different letters.When the vowels EAI are always together, they can be supposed to form oneletter.Then, we have to arrange the letters LNDG (EAI).Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.The vowels (EAI) can be arranged among themselves in 3! = 6 ways.Required number of ways = (120 x 6) = 720.4. In how many ways can the letters of the word LEADER be arranged?A.72 B.144 C.360 D.720 E.None of theseAnswer: Option CExplanation:The word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.Required number of ways = = 360)!1)(!1)(!1)(!2)(!1(!6
- 9. Thank you!!

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