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Permutation and CombinationQuestion and Answers: Page 11

Question Number 123577 Answers: 1 Comments: 0

Question Number 122250 Answers: 1 Comments: 0

Question Number 121829 Answers: 1 Comments: 0

Question Number 121851 Answers: 0 Comments: 1

How many ways to arrange letters using letters in the word OKINKO (a) 1260 (b) 1160 (c) 980 (d) 880 (e) 720

$${How}\:{many}\:{ways}\:{to}\:{arrange}\:{letters}\: \\ $$$${using}\:{letters}\:{in}\:{the}\:{word}\:{OKINKO} \\ $$$$\left({a}\right)\:\mathrm{1260}\:\:\:\:\:\left({b}\right)\:\mathrm{1160}\:\:\:\:\left({c}\right)\:\mathrm{980} \\ $$$$\left({d}\right)\:\mathrm{880}\:\:\:\:\:\:\:\:\left({e}\right)\:\mathrm{720} \\ $$

Question Number 120954 Answers: 0 Comments: 0

shown that n!=Σ_(k=0) ^n (−1)^k (_( k) ^( n) )(n−k)^n please help me

$$\boldsymbol{{shown}}\:\boldsymbol{{that}} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{n}}!=\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\left(−\mathrm{1}\right)^{\boldsymbol{{k}}} \left(_{\:\boldsymbol{{k}}} ^{\:\boldsymbol{{n}}} \right)\left(\boldsymbol{{n}}−\boldsymbol{{k}}\right)^{\boldsymbol{{n}}} \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}}\:\boldsymbol{{me}} \\ $$

Question Number 120431 Answers: 1 Comments: 2

Question Number 120209 Answers: 1 Comments: 0

Peter has 12 relatives (5 man & 7 woman) and his wife also has 12 relatives (5 woman &7 man). They do not have common relatives. They decided to invite 12 guests ,six each of their relatives, such that there are six man and six woman among the guests. How many ways can they choose 12 guests?

$${Peter}\:{has}\:\mathrm{12}\:{relatives}\:\left(\mathrm{5}\:{man}\:\&\:\mathrm{7}\:{woman}\right) \\ $$$${and}\:{his}\:{wife}\:{also}\:{has}\:\mathrm{12}\:{relatives} \\ $$$$\left(\mathrm{5}\:{woman}\:\&\mathrm{7}\:{man}\right).\:{They}\:{do}\:{not} \\ $$$${have}\:{common}\:{relatives}.\:{They}\:{decided} \\ $$$${to}\:{invite}\:\mathrm{12}\:{guests}\:,{six}\:{each}\:{of} \\ $$$${their}\:{relatives},\:{such}\:{that}\:{there} \\ $$$${are}\:{six}\:{man}\:{and}\:{six}\:{woman}\: \\ $$$${among}\:{the}\:{guests}.\:{How}\:{many} \\ $$$${ways}\:{can}\:{they}\:{choose}\:\mathrm{12}\:{guests}? \\ $$

Question Number 120188 Answers: 0 Comments: 0

Let n be a positive integer . Prove that Σ_(k=0) ^n 2^k ((n),(k) ) ((( n−k)),((⌊((n−k)/2)⌋)) ) = (((2n+1)),(( n)) )

$${Let}\:{n}\:{be}\:{a}\:{positive}\:{integer}\:. \\ $$$${Prove}\:{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:\begin{pmatrix}{\:\:{n}−{k}}\\{\lfloor\frac{{n}−{k}}{\mathrm{2}}\rfloor}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:{n}}\end{pmatrix} \\ $$

Question Number 120090 Answers: 0 Comments: 0

Find the number of subsets of { 1,2,3,...,2000 } the sum of whose elements is divisible by 5

$${Find}\:{the}\:{number}\:{of}\:{subsets}\:{of} \\ $$$$\left\{\:\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{2000}\:\right\}\:{the}\:{sum}\:{of}\: \\ $$$${whose}\:{elements}\:{is}\:{divisible}\:{by}\:\mathrm{5} \\ $$

Question Number 119896 Answers: 2 Comments: 1

Question Number 119747 Answers: 1 Comments: 0

Suppose that 7 blue balls , 8 red balls and 9 green balls should be put into three boxes labeled 1,2 and 3, so that any box contains at least one balls of each colour. How many ways can this arrangement be done?

$${Suppose}\:{that}\:\mathrm{7}\:{blue}\:{balls}\:,\:\mathrm{8}\:{red}\:{balls}\:{and}\:\mathrm{9}\:{green} \\ $$$${balls}\:{should}\:{be}\:{put}\:{into}\:{three}\:{boxes}\:{labeled} \\ $$$$\mathrm{1},\mathrm{2}\:{and}\:\mathrm{3},\:{so}\:{that}\:{any}\:{box}\:{contains}\:{at}\:{least} \\ $$$${one}\:{balls}\:{of}\:{each}\:{colour}.\:{How}\:{many}\:{ways} \\ $$$${can}\:{this}\:{arrangement}\:{be}\:{done}? \\ $$

Question Number 119733 Answers: 1 Comments: 5

Question Number 119479 Answers: 3 Comments: 0

Π_(k=1) ^∞ cos((x/2^k )) = ?

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:=\:? \\ $$

Question Number 119397 Answers: 2 Comments: 1

Find the number of possible arrangements of the letters in the word PENCILS if (a) ′E′ is next to ′I′ (b) E comes before I (c) there are three letters between E and I

$${Find}\:{the}\:{number}\:{of}\:{possible}\:{arrangements} \\ $$$${of}\:{the}\:{letters}\:{in}\:{the}\:{word}\:{PENCILS}\:{if} \\ $$$$\left({a}\right)\:'{E}'\:{is}\:{next}\:{to}\:'{I}' \\ $$$$\left({b}\right)\:{E}\:{comes}\:{before}\:{I} \\ $$$$\left({c}\right)\:{there}\:{are}\:{three}\:{letters}\:{between}\:{E}\:{and}\:{I} \\ $$

Question Number 119390 Answers: 2 Comments: 0

we have 15 different mathematics books, 10 different physics books and 12 different chemistry books. we should choose 6 books such that they contain all three kinds of books. in how many ways can we do this?

$${we}\:{have}\:\mathrm{15}\:{different}\:{mathematics} \\ $$$${books},\:\mathrm{10}\:{different}\:{physics}\:{books}\:{and} \\ $$$$\mathrm{12}\:{different}\:{chemistry}\:{books}.\:{we}\:{should} \\ $$$${choose}\:\mathrm{6}\:{books}\:{such}\:{that}\:{they}\:{contain} \\ $$$${all}\:{three}\:{kinds}\:{of}\:{books}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{we}\:{do}\:{this}? \\ $$

Question Number 119364 Answers: 2 Comments: 0

Suppose once more we′re asked to choose four students from high school class of 15 to form a committee but this time we have a restriction : we don′t want to committee to consist of all seniors or all juniors .suppose there are eight seniors and seven juniors in the class. How many different committe can we form?

$${Suppose}\:{once}\:{more}\:{we}'{re}\:{asked}\: \\ $$$${to}\:{choose}\:{four}\:{students}\:{from}\:{high}\:{school} \\ $$$${class}\:{of}\:\mathrm{15}\:{to}\:{form}\:{a}\:{committee} \\ $$$${but}\:{this}\:{time}\:{we}\:{have}\:{a}\:{restriction} \\ $$$$:\:{we}\:{don}'{t}\:{want}\:{to}\:{committee}\:{to} \\ $$$${consist}\:{of}\:{all}\:{seniors}\:{or}\:{all}\:{juniors} \\ $$$$.{suppose}\:{there}\:{are}\:{eight}\:{seniors} \\ $$$${and}\:{seven}\:{juniors}\:{in}\:{the}\:{class}.\:{How}\:{many} \\ $$$${different}\:{committe}\:{can}\:{we}\:{form}? \\ $$$$ \\ $$

Question Number 118822 Answers: 2 Comments: 0

In how many ways can the letters of the word LEVITATE be arranged if the vowels must not be next to each other

$$\:\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\: \\ $$$$\:\mathrm{the}\:\mathrm{word}\:{LEVITATE}\:\mathrm{be}\:\mathrm{arranged}\:\mathrm{if} \\ $$$$\:\mathrm{the}\:\mathrm{vowels}\:\mathrm{must}\:\mathrm{not}\:\mathrm{be}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each} \\ $$$$\:\mathrm{other} \\ $$

Question Number 118694 Answers: 2 Comments: 0

How many sets of 3 numbers each can be formed from the numbers { 1,2,3,...,20 } if no two consecutive numbers are to be in a set ?

$${How}\:{many}\:{sets}\:{of}\:\mathrm{3}\:{numbers}\:{each}\:{can} \\ $$$${be}\:{formed}\:{from}\:{the}\:{numbers}\:\left\{\:\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{20}\:\right\}\:{if}\:{no} \\ $$$${two}\:{consecutive}\:{numbers}\:{are}\:{to}\:{be}\:{in}\:{a}\:{set}\:? \\ $$

Question Number 118555 Answers: 2 Comments: 0

Question Number 118489 Answers: 2 Comments: 0

(1) solve the equation ((x−49)/(50)) + ((x−50)/(49)) = ((49)/(x−50)) + ((50)/(x−49)) (2) How many numbers from 12 to 12345 inclusive have digits which are consecutive an in increasing order, reading from left to right ?

$$\left(\mathrm{1}\right)\:{solve}\:{the}\:{equation}\:\frac{{x}−\mathrm{49}}{\mathrm{50}}\:+\:\frac{{x}−\mathrm{50}}{\mathrm{49}}\:=\:\frac{\mathrm{49}}{{x}−\mathrm{50}}\:+\:\frac{\mathrm{50}}{{x}−\mathrm{49}} \\ $$$$\left(\mathrm{2}\right)\:{How}\:{many}\:{numbers}\:{from}\:\mathrm{12}\:{to}\:\mathrm{12345}\: \\ $$$${inclusive}\:{have}\:{digits}\:{which}\:{are}\: \\ $$$${consecutive}\:{an}\:{in}\:{increasing}\:{order}, \\ $$$${reading}\:{from}\:{left}\:{to}\:{right}\:?\: \\ $$

Question Number 117708 Answers: 2 Comments: 3

Find the number of all 5 digit numbers x_1 x_2 x_3 x_4 x_5 with x_1 ≥x_2 ≥x_3 ≥x_4 ≥x_5 .

$${Find}\:{the}\:{number}\:{of}\:{all}\:\mathrm{5}\:{digit} \\ $$$${numbers}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} {x}_{\mathrm{4}} {x}_{\mathrm{5}} \:{with} \\ $$$${x}_{\mathrm{1}} \geqslant{x}_{\mathrm{2}} \geqslant{x}_{\mathrm{3}} \geqslant{x}_{\mathrm{4}} \geqslant{x}_{\mathrm{5}} . \\ $$

Question Number 117281 Answers: 1 Comments: 0

In how many different ways can we select 5 numbers from 9 numbers {1,2,3,...,9}? A number may be selected more than one time.

$${In}\:{how}\:{many}\:{different}\:{ways}\:{can}\:{we} \\ $$$${select}\:\mathrm{5}\:{numbers}\:{from}\:\mathrm{9}\:{numbers} \\ $$$$\left\{\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{9}\right\}?\: \\ $$$${A}\:{number}\:{may}\:{be}\:{selected}\:{more}\:{than} \\ $$$${one}\:{time}. \\ $$

Question Number 117192 Answers: 1 Comments: 0

Assuming you have enough coins of 1,5,10,25,and 50cents. In how many ways can you make a change for 1dollar.

$$\mathrm{Assuming}\:\mathrm{you}\:\mathrm{have}\:\mathrm{enough}\:\mathrm{coins}\: \\ $$$$\mathrm{of}\:\mathrm{1},\mathrm{5},\mathrm{10},\mathrm{25},\mathrm{and}\:\mathrm{50cents}.\:\mathrm{In}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{you}\:\mathrm{make}\:\mathrm{a}\:\mathrm{change}\:\mathrm{for}\:\mathrm{1dollar}. \\ $$

Question Number 116883 Answers: 1 Comments: 0

Find the number m of ways to partition 10 students into four team so that two team contains 3 students and two team contains 2 students .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{m}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to} \\ $$$$\mathrm{partition}\:\mathrm{10}\:\mathrm{students}\:\mathrm{into}\:\mathrm{four}\:\mathrm{team} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{two}\:\mathrm{team}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{students} \\ $$$$\mathrm{and}\:\mathrm{two}\:\mathrm{team}\:\mathrm{contains}\:\mathrm{2}\:\mathrm{students}\:. \\ $$

Question Number 116881 Answers: 3 Comments: 0

Find the number m of non negative integer solution of x+y+z=18

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{m}\:\mathrm{of}\:\mathrm{non}\:\mathrm{negative} \\ $$$$\mathrm{integer}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{18} \\ $$$$ \\ $$

Question Number 116397 Answers: 2 Comments: 0

The number 1,2,3,4,...,7 are randomly divided into two non −empty subsets. The probability that the sum of the numbers in the two subsets being equal is (r/s) expressed in the lowest term. Find r+s ?

$${The}\:{number}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},...,\mathrm{7}\:{are}\:{randomly} \\ $$$${divided}\:{into}\:{two}\:{non}\:−{empty}\:{subsets}. \\ $$$${The}\:{probability}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${numbers}\:{in}\:{the}\:{two}\:{subsets}\:{being} \\ $$$${equal}\:{is}\:\frac{{r}}{{s}}\:{expressed}\:{in}\:{the}\:{lowest} \\ $$$${term}.\:{Find}\:{r}+{s}\:? \\ $$

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