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Permutation and CombinationQuestion and Answers: Page 10

Question Number 131744 Answers: 1 Comments: 0

Question Number 131635 Answers: 0 Comments: 0

three subject group are to be formed randomly by 15 students (including 3 girls) under the condition that each groups consist 5 students and each student attends only one group. flnd the probabilities that of the following events (1) there is exactly one girl in each group (2) the 3 girls attend the same group

$$ \\ $$$$\mathrm{three}\:\mathrm{subject}\:\mathrm{group}\:\mathrm{are}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{formed}\:\mathrm{randomly}\:\mathrm{by}\:\mathrm{15}\:\mathrm{students} \\ $$$$\left(\mathrm{including}\:\mathrm{3}\:\mathrm{girls}\right)\:\mathrm{under}\:\mathrm{the} \\ $$$$\mathrm{condition}\:\mathrm{that}\:\mathrm{each}\:\mathrm{groups} \\ $$$$\mathrm{consist}\:\mathrm{5}\:\mathrm{students}\:\mathrm{and}\:\mathrm{each} \\ $$$$\mathrm{student}\:\mathrm{attends}\:\mathrm{only}\:\mathrm{one}\:\mathrm{group}. \\ $$$$\mathrm{flnd}\:\mathrm{the}\:\mathrm{probabilities}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{events}\:\left(\mathrm{1}\right)\:\mathrm{there}\:\mathrm{is} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{girl}\:\mathrm{in}\:\mathrm{each}\:\mathrm{group}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{the}\:\mathrm{3}\:\mathrm{girls}\:\mathrm{attend}\:\mathrm{the}\:\mathrm{same}\:\mathrm{group} \\ $$

Question Number 131553 Answers: 1 Comments: 0

Question Number 131464 Answers: 1 Comments: 0

determinant (((old plate : EEU 874)),((new plate : 1BXK 267))) Old California license plate consisted of a sequence of three letters followed by three digits (see figure above). Assuming that any sequence of letters and digits was allowed (though actually some combinations of letters were disallowed), how many license plate were available ?

$$\:\begin{array}{|c|c|}{{old}\:{plate}\::\:{EEU}\:\mathrm{874}}\\{{new}\:{plate}\::\:\mathrm{1}{BXK}\:\mathrm{267}}\\\hline\end{array} \\ $$$${Old}\:{California}\:{license}\:{plate}\: \\ $$$${consisted}\:{of}\:{a}\:{sequence}\:{of} \\ $$$${three}\:{letters}\:{followed}\:{by}\:{three} \\ $$$${digits}\:\left({see}\:{figure}\:{above}\right). \\ $$$${Assuming}\:{that}\:{any}\:{sequence} \\ $$$${of}\:{letters}\:{and}\:{digits}\:{was}\:{allowed} \\ $$$$\left({though}\:{actually}\:{some}\:{combinations}\right. \\ $$$$\left.{of}\:{letters}\:{were}\:{disallowed}\right),\:{how} \\ $$$${many}\:{license}\:{plate}\:{were}\: \\ $$$${available}\:? \\ $$

Question Number 131459 Answers: 1 Comments: 0

Eight eligible bachelor and seven beautiful models happen randomly to have purchased single seats in the same 15−seats row of theather. On the average , how many pairs of adjacent seats are ticketed for marriageable couples ?

$${Eight}\:{eligible}\:{bachelor}\:{and}\:{seven}\:{beautiful} \\ $$$${models}\:{happen}\:{randomly}\:{to}\:{have}\: \\ $$$${purchased}\:{single}\:{seats}\:{in}\:{the}\:{same}\:\mathrm{15}−{seats} \\ $$$${row}\:{of}\:{theather}.\:{On}\:{the}\:{average}\:,\:{how}\:{many} \\ $$$${pairs}\:{of}\:{adjacent}\:{seats}\:{are}\:{ticketed} \\ $$$${for}\:{marriageable}\:{couples}\:? \\ $$

Question Number 131248 Answers: 2 Comments: 0

If α and β are the coefficient of x^8 and x^(−24) respectively in the expansion of [ x^4 +2+(1/x^4 ) ]^(10) in powers of x then (α/β) is equal to

$${If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{coefficient}\: \\ $$$${of}\:{x}^{\mathrm{8}} \:{and}\:{x}^{−\mathrm{24}} \:{respectively}\: \\ $$$${in}\:{the}\:{expansion}\:{of}\:\left[\:{x}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\right]^{\mathrm{10}} \\ $$$${in}\:{powers}\:{of}\:{x}\:{then}\:\frac{\alpha}{\beta}\:{is}\:{equal}\:{to}\: \\ $$

Question Number 131162 Answers: 0 Comments: 2

Five children sitting one behind the other in a five seater merry−go−round ,decide to switch seats so that each child has new companion in front. In how many ways can this be done?

$${Five}\:{children}\:{sitting}\:{one}\:{behind} \\ $$$${the}\:{other}\:{in}\:{a}\:{five}\:{seater}\:{merry}−{go}−{round} \\ $$$$,{decide}\:{to}\:{switch}\:{seats}\:{so}\:{that}\:{each} \\ $$$${child}\:{has}\:{new}\:{companion}\:{in}\:{front}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{this}\:{be}\:{done}? \\ $$

Question Number 130899 Answers: 0 Comments: 1

There are 20 persons at a party. Maria dansed with 7 boys, Stacy dansed with 8 boys, Monia dansed with 9 boys and Eve dansed with all the boys at the party. How many girls are there at the party ?

$$\mathrm{There}\:\mathrm{are}\:\mathrm{20}\:\mathrm{persons}\:\mathrm{at}\:\mathrm{a}\:\mathrm{party}. \\ $$$$\mathrm{Maria}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{7}\:\mathrm{boys}, \\ $$$$\mathrm{Stacy}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{8}\:\mathrm{boys}, \\ $$$$\mathrm{Monia}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{9}\:\mathrm{boys}\:\mathrm{and} \\ $$$$\mathrm{Eve}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{all}\:\mathrm{the}\:\mathrm{boys}\:\mathrm{at}\:\mathrm{the}\:\mathrm{party}. \\ $$$$\boldsymbol{\mathrm{How}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{girls}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{there}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{party}}\:? \\ $$

Question Number 130685 Answers: 2 Comments: 0

How many password containing 6 characters of the letters in the word ′′ Move Now ′′ ?

$${How}\:{many}\:{password}\:{containing} \\ $$$$\mathrm{6}\:{characters}\:{of}\:{the}\:{letters}\:{in}\:{the}\: \\ $$$${word}\:''\:{Move}\:{Now}\:''\:? \\ $$

Question Number 130689 Answers: 0 Comments: 0

Mr. Rahmat decided to create a password with the form numbers and letterse intermittntly intermittent ( can also be letters and numbers intermittently) but no nmbers and letters are the same. It chooses to use numbers on the set {2, 5, 8, 9} and selects the letters on the set { A, X, Y, W,Z} How many passwords can I create

$$ \\ $$$$\mathrm{Mr}.\:\mathrm{Rahmat}\:\mathrm{decided}\:\mathrm{to}\:\mathrm{create}\:\mathrm{a}\:\mathrm{password}\: \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{form}\:\mathrm{numbers}\:\mathrm{and}\:\mathrm{letterse} \\ $$$$\mathrm{intermittntly}\:\:\mathrm{intermittent}\:\left(\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\right. \\ $$$$\left.\mathrm{letters}\:\mathrm{and}\:\mathrm{numbers}\:\mathrm{intermittently}\right)\:\mathrm{but}\:\mathrm{no} \\ $$$$\mathrm{nmbers}\:\mathrm{and}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same}.\:\mathrm{It}\: \\ $$$$\mathrm{chooses}\:\mathrm{to}\:\mathrm{use}\:\mathrm{numbers}\:\mathrm{on}\:\mathrm{the}\:\mathrm{set}\:\:\left\{\mathrm{2},\:\mathrm{5},\:\mathrm{8},\:\mathrm{9}\right\}\: \\ $$$$\mathrm{and}\:\mathrm{selects}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{on}\:\mathrm{the}\:\mathrm{set}\:\left\{\:\mathrm{A},\:\mathrm{X},\:\mathrm{Y},\:\mathrm{W},\mathrm{Z}\right\}\: \\ $$$$\:\mathrm{How}\:\mathrm{many}\:\mathrm{passwords}\:\mathrm{can}\:\mathrm{I}\:\mathrm{create}\: \\ $$

Question Number 130634 Answers: 1 Comments: 0

Question Number 130602 Answers: 1 Comments: 0

Question Number 130193 Answers: 2 Comments: 0

How many words with at least 2 letters can be formed from UNUSUALNESS?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{words}\:\mathrm{with}\:\mathrm{at}\:\mathrm{least}\:\mathrm{2}\:\mathrm{letters} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\boldsymbol{\mathrm{UNUSUALNESS}}? \\ $$

Question Number 130034 Answers: 1 Comments: 0

How many integers between 999 and 4000 (both are included) can be formed using digits 0,1,2,3,4 if repetition of digits is allowed?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{integers}\:\mathrm{between}\:\mathrm{999}\:\mathrm{and}\:\mathrm{4000} \\ $$$$\left(\mathrm{both}\:\mathrm{are}\:\mathrm{included}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{using} \\ $$$$\mathrm{digits}\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\mathrm{if}\:\mathrm{repetition}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{is} \\ $$$$\mathrm{allowed}?\: \\ $$

Question Number 130019 Answers: 1 Comments: 0

How many 4 letters word can be formed from COMMUNICATION

$$\mathrm{How}\:\mathrm{many}\:\mathrm{4}\:\mathrm{letters}\:\mathrm{word}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\:\:\:\mathrm{COMMUNICATION} \\ $$

Question Number 125548 Answers: 1 Comments: 2

How many four−digit numbers between 2000 and 4000 can be formed from the numbers 0, 1, 2, 3, 4 and 5 if repitition is allowed?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{four}−\mathrm{digit}\:\mathrm{numbers}\: \\ $$$$\mathrm{between}\:\mathrm{2000}\:\mathrm{and}\:\mathrm{4000}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\mathrm{and}\:\mathrm{5}\: \\ $$$$\mathrm{if}\:\mathrm{repitition}\:\mathrm{is}\:\mathrm{allowed}? \\ $$

Question Number 125460 Answers: 1 Comments: 0

In a room containing N people N>3, at least one person has not shaken hands with everyone else in the room. What is the maximum number of people in the room that could have shaken hands with everyone else?

$$\:{In}\:{a}\:{room}\:{containing}\:{N}\:{people} \\ $$$${N}>\mathrm{3},\:{at}\:{least}\:{one}\:{person}\:{has}\:{not} \\ $$$${shaken}\:{hands}\:{with}\:{everyone} \\ $$$${else}\:{in}\:{the}\:{room}.\:{What}\:{is}\:{the}\: \\ $$$${maximum}\:{number}\:{of}\:{people}\: \\ $$$${in}\:{the}\:{room}\:{that}\:{could}\:{have}\: \\ $$$${shaken}\:{hands}\:{with}\:{everyone}\: \\ $$$${else}?\: \\ $$

Question Number 125208 Answers: 1 Comments: 0

There are 12 students in a party. Five of them are girls. In how many ways can these 12 students be arranged in a row if (i) there are no restrictions? (ii) the 5 girls must be together (forming a block)? (iii) no 2 girls are adjacent (iv) between two particular boys A and B , there no boys but exactly 3 girls?

$${There}\:{are}\:\mathrm{12}\:{students}\:{in}\:{a}\:{party}.\:{Five}\:{of} \\ $$$${them}\:{are}\:{girls}.\:{In}\:{how}\:{many}\:{ways}\:{can}\: \\ $$$${these}\:\mathrm{12}\:{students}\:{be}\:{arranged}\:{in}\:{a}\:{row}\:{if}\: \\ $$$$\left({i}\right)\:{there}\:{are}\:{no}\:{restrictions}? \\ $$$$\left({ii}\right)\:{the}\:\mathrm{5}\:{girls}\:{must}\:{be}\:{together}\:\left({forming}\:{a}\:{block}\right)? \\ $$$$\left({iii}\right)\:{no}\:\mathrm{2}\:{girls}\:{are}\:{adjacent}\: \\ $$$$\left({iv}\right)\:{between}\:{two}\:{particular}\:{boys}\:{A}\:{and}\:{B}\: \\ $$$$\:,\:{there}\:{no}\:{boys}\:{but}\:{exactly}\:\mathrm{3}\:{girls}? \\ $$

Question Number 125207 Answers: 2 Comments: 0

Find the number of ways to choose a pair {a,b} of distinct numbers from the set {1,2,3,...,50} such that (i) ∣a−b∣ = 5 (ii) ∣a−b∣ ≤ 5

$${Find}\:{the}\:{number}\:{of}\:{ways}\:{to}\:{choose}\:{a}\:{pair} \\ $$$$\left\{{a},{b}\right\}\:{of}\:{distinct}\:{numbers}\:{from}\:{the}\:{set}\: \\ $$$$\left\{\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{50}\right\}\:{such}\:{that}\: \\ $$$$\left({i}\right)\:\mid{a}−{b}\mid\:=\:\mathrm{5} \\ $$$$\left({ii}\right)\:\mid{a}−{b}\mid\:\leqslant\:\mathrm{5}\: \\ $$

Question Number 125191 Answers: 0 Comments: 4

In how many ways can 3 men and 5 women be seated in a row if 3 specific women cannot sit next to each other? My solution: arrangements when the 3 women cannot sit next to eachother = 5! ×^6 P_3 = 14,400 Solution in book: arrangements with no restrictions = 8! arrangements with the 3 women seated next to eachother = 3! × 6! arrangements when the 3 women cannot sit next to eachother = 8! − (3! × 6!) = 36,000 I′m confused. Can I please get some explanation why my solution is wrong?

$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{3}\:\mathrm{men}\:\mathrm{and}\:\mathrm{5} \\ $$$$\mathrm{women}\:\mathrm{be}\:\mathrm{seated}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row}\:\mathrm{if}\:\mathrm{3}\:\mathrm{specific} \\ $$$$\mathrm{women}\:{cannot}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}? \\ $$$$ \\ $$$$\mathrm{My}\:\mathrm{solution}: \\ $$$$\mathrm{arrangements}\:\mathrm{when}\:\mathrm{the}\:\mathrm{3}\:\mathrm{women} \\ $$$${cannot}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{5}!\:×\:^{\mathrm{6}} {P}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{14},\mathrm{400} \\ $$$$ \\ $$$$\mathrm{Solution}\:\mathrm{in}\:\mathrm{book}: \\ $$$$\mathrm{arrangements}\:\mathrm{with}\:\mathrm{no}\:\mathrm{restrictions}\:=\:\mathrm{8}! \\ $$$$ \\ $$$$\mathrm{arrangements}\:\mathrm{with}\:\mathrm{the}\:\mathrm{3}\:\mathrm{women} \\ $$$$\mathrm{seated}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}\:\:=\:\:\mathrm{3}!\:×\:\mathrm{6}! \\ $$$$ \\ $$$$\mathrm{arrangements}\:\mathrm{when}\:\mathrm{the}\:\mathrm{3}\:\mathrm{women} \\ $$$${cannot}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{8}!\:−\:\left(\mathrm{3}!\:×\:\mathrm{6}!\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{36},\mathrm{000} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{confused}.\:\mathrm{Can}\:\mathrm{I}\:\mathrm{please}\:\mathrm{get}\:\mathrm{some} \\ $$$$\mathrm{explanation}\:\mathrm{why}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{wrong}? \\ $$

Question Number 124916 Answers: 2 Comments: 0

The number of ways arrangements of the word ′MASKARA′ with exactly 2 A′s are adjacent??

$${The}\:{number}\:{of}\:{ways}\:{arrangements}\: \\ $$$${of}\:{the}\:{word}\:'{MASKARA}'\:{with}\:{exactly} \\ $$$$\mathrm{2}\:{A}'{s}\:\:{are}\:{adjacent}??\: \\ $$

Question Number 124878 Answers: 3 Comments: 0

20 students should stand in 5 different rows. each row should have at least 2 students. in how many ways can you arrange them?

$$\mathrm{20}\:{students}\:{should}\:{stand}\:{in}\:\mathrm{5} \\ $$$${different}\:{rows}.\:{each}\:{row}\:{should}\:{have} \\ $$$${at}\:{least}\:\mathrm{2}\:{students}.\:{in}\:{how}\:{many}\:{ways} \\ $$$${can}\:{you}\:{arrange}\:{them}? \\ $$

Question Number 124829 Answers: 2 Comments: 0

How many ways are there to arrange the letters of the word ′ALAMATAR′ if no two A′s are adjacent?

$$\:{How}\:{many}\:{ways}\:{are}\:{there}\:{to}\:{arrange} \\ $$$${the}\:{letters}\:{of}\:{the}\:{word}\:'{ALAMATAR}'\:{if} \\ $$$${no}\:{two}\:{A}'{s}\:{are}\:{adjacent}?\: \\ $$

Question Number 124826 Answers: 2 Comments: 0

How many ways are there to arrange the letters of the word ′ VISITING′ if no two I′s are adjacent ?

$$\:{How}\:{many}\:{ways}\:{are}\:{there}\:{to}\:{arrange}\: \\ $$$${the}\:{letters}\:{of}\:{the}\:{word}\:'\:{VISITING}' \\ $$$${if}\:{no}\:{two}\:{I}'{s}\:{are}\:{adjacent}\:? \\ $$

Question Number 124806 Answers: 2 Comments: 0

Question Number 124712 Answers: 2 Comments: 0

In how many ways can 5 boys and 3 girls be seated around a table if (i) boy B_3 and G_2 are not adjacent (ii) no girls are adjacent

$${In}\:{how}\:{many}\:{ways}\:\:{can}\:\mathrm{5}\:{boys}\:{and}\:\mathrm{3}\:{girls} \\ $$$${be}\:{seated}\:{around}\:{a}\:{table}\:{if}\: \\ $$$$\left({i}\right)\:{boy}\:{B}_{\mathrm{3}} \:{and}\:{G}_{\mathrm{2}} \:{are}\:{not}\:{adjacent} \\ $$$$\left({ii}\right)\:{no}\:{girls}\:{are}\:{adjacent}\: \\ $$

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