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Permutation and CombinationQuestion and Answers: Page 1

Question Number 218578 Answers: 2 Comments: 0

10 couples are invited to a dinner. after the dinner they form pairs to dance. in how many ways can they do this, 1) generally 2) a man should dance with a woman 3) as 2), but two special couples should not dance with each other 4) a man should dance with a woman, but not with his own wife 5) as 4), but two special couples should not dance with each other

$$\mathrm{10}\:{couples}\:{are}\:{invited}\:{to}\:{a}\:{dinner}. \\ $$$${after}\:{the}\:{dinner}\:{they}\:{form}\:{pairs}\: \\ $$$${to}\:{dance}.\:{in}\:{how}\:{many}\:{ways}\:{can}\:{they} \\ $$$${do}\:{this}, \\ $$$$\left.\mathrm{1}\right)\:{generally} \\ $$$$\left.\mathrm{2}\right)\:{a}\:{man}\:{should}\:{dance}\:{with}\:{a}\:{woman} \\ $$$$\left.\mathrm{3}\left.\right)\:{as}\:\mathrm{2}\right),\:{but}\:{two}\:{special}\:{couples} \\ $$$$\:\:\:\:\:{should}\:{not}\:{dance}\:{with}\:{each}\:{other} \\ $$$$\left.\mathrm{4}\right)\:{a}\:{man}\:{should}\:{dance}\:{with}\:{a}\:{woman}, \\ $$$$\:\:\:\:\:{but}\:{not}\:{with}\:{his}\:{own}\:{wife} \\ $$$$\left.\mathrm{5}\left.\right)\:{as}\:\mathrm{4}\right),\:{but}\:{two}\:{special}\:{couples} \\ $$$$\:\:\:\:\:{should}\:{not}\:{dance}\:{with}\:{each}\:{other} \\ $$

Question Number 218445 Answers: 3 Comments: 0

there are 100 students in a school. it is found out that each student should select at least 4 courses, so that no two students have the same selection. how many different courses does the school offer?

$${there}\:{are}\:\mathrm{100}\:{students}\:{in}\:{a}\:{school}. \\ $$$${it}\:{is}\:{found}\:{out}\:{that}\:{each}\:{student}\: \\ $$$${should}\:{select}\:{at}\:{least}\:\mathrm{4}\:{courses},\:{so}\: \\ $$$${that}\:{no}\:{two}\:{students}\:{have}\:{the}\:{same}\: \\ $$$${selection}.\: \\ $$$${how}\:{many}\:{different}\:{courses}\:{does}\: \\ $$$${the}\:{school}\:{offer}? \\ $$

Question Number 218310 Answers: 2 Comments: 1

10 couples are invited to a dinner and should be seated at a round table. in how many ways can the host do this, 1) generally. 2) if the husband and the wife of each couple should sit together. 3) if no two men should sit next to each other. 4) as 3), but two speicial couples should not sit next to each other. it means that the husband from couple 1 may not sit next to the wife from couple 2 and the husband from couple 2 may not sit next to the wife from couple 1. but certainly the husbands of both couples may sit next to their own wifes.

$$\mathrm{10}\:{couples}\:{are}\:{invited}\:{to}\:{a}\:{dinner} \\ $$$${and}\:{should}\:{be}\:{seated}\:{at}\:{a}\:{round}\:{table}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{host}\:{do} \\ $$$${this}, \\ $$$$\left.\mathrm{1}\right)\:{generally}. \\ $$$$\left.\mathrm{2}\right)\:{if}\:{the}\:{husband}\:{and}\:{the}\:{wife}\:{of} \\ $$$$\:\:\:\:\:{each}\:{couple}\:{should}\:{sit}\:{together}. \\ $$$$\left.\mathrm{3}\right)\:{if}\:{no}\:{two}\:{men}\:{should}\:{sit}\:{next} \\ $$$$\:\:\:\:\:{to}\:{each}\:{other}. \\ $$$$\left.\mathrm{4}\left.\right)\:{as}\:\mathrm{3}\right),\:{but}\:{two}\:{speicial}\:{couples}\: \\ $$$$\:\:\:\:\:{should}\:{not}\:{sit}\:{next}\:{to}\:{each}\:{other}. \\ $$$$\:\:\:\:\underline{\:{it}\:{means}\:}{that}\:{the}\:{husband}\:{from}\: \\ $$$$\:\:\:\:\:{couple}\:\mathrm{1}\:{may}\:{not}\:{sit}\:{next}\:{to}\:{the} \\ $$$$\:\:\:\:\:{wife}\:{from}\:{couple}\:\mathrm{2}\:{and}\:{the} \\ $$$$\:\:\:\:\:\:{husband}\:{from}\:{couple}\:\mathrm{2}\:{may}\:{not} \\ $$$$\:\:\:\:\:\:{sit}\:{next}\:{to}\:{the}\:{wife}\:{from}\:{couple}\:\mathrm{1}. \\ $$$$\:\:\:\:\:\:{but}\:{certainly}\:{the}\:{husbands}\:{of} \\ $$$$\:\:\:\:\:\:{both}\:{couples}\:{may}\:{sit}\:{next}\:{to}\:{their} \\ $$$$\:\:\:\:\:\:{own}\:{wifes}. \\ $$

Question Number 218169 Answers: 0 Comments: 0

P(5,6)=((15!)/((15−6))) = ((15!)/(9!)) =((15×14×131×2×11×10×9×8×7×6×5×4×3×2×1)/(9×8×7×6×5×4×3×2×)) =15×14×13×12×11×10 =3,603,600

$${P}\left(\mathrm{5},\mathrm{6}\right)=\frac{\mathrm{15}!}{\left(\mathrm{15}−\mathrm{6}\right)}\:=\:\frac{\mathrm{15}!}{\mathrm{9}!}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{15}×\mathrm{14}×\mathrm{131}×\mathrm{2}×\mathrm{11}×\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{15}×\mathrm{14}×\mathrm{13}×\mathrm{12}×\mathrm{11}×\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3},\mathrm{603},\mathrm{600} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 218129 Answers: 2 Comments: 0

how many different words can be formed from the word MATHEMATICS? note: here a word should have at least two letters, but mustn′t have a meaning.

$${how}\:{many}\:{different}\:{words}\:{can}\:{be} \\ $$$${formed}\:{from}\:{the}\:{word}\: \\ $$$$\boldsymbol{\mathrm{MATHEMATICS}}? \\ $$$${note}:\:\:{here}\:{a}\:{word}\:{should}\:{have}\:{at}\: \\ $$$${least}\:{two}\:{letters},\:{but}\:{mustn}'{t}\:{have}\:{a} \\ $$$${meaning}. \\ $$

Question Number 217402 Answers: 3 Comments: 0

Question Number 214923 Answers: 0 Comments: 0

Question Number 213796 Answers: 4 Comments: 0

Question Number 212686 Answers: 4 Comments: 4

in how many ways can a teacher divide his 10 studens into 4 groups such that each group has at least 2 students?

$${in}\:{how}\:{many}\:{ways}\:{can}\:{a}\:{teacher} \\ $$$${divide}\:{his}\:\mathrm{10}\:{studens}\:{into}\:\mathrm{4}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{has}\:{at}\:{least}\:\mathrm{2}\: \\ $$$${students}? \\ $$

Question Number 210290 Answers: 1 Comments: 0

Question Number 209281 Answers: 2 Comments: 2

Find the value of r, if ^(10) C_r = ^(10) C_(2r + 1)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{r},\:\mathrm{if}\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{r}} \:\:=\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{2r}\:\:+\:\:\mathrm{1}} \\ $$

Question Number 208662 Answers: 2 Comments: 0

(( ((n),(0) ) +3 ((n),(1) ) +5 ((n),(2) ) +...+(2n+1) ((n),(n) ))/( ((n),(1) ) +2 ((n),(2) ) + 3 ((n),(3) ) +...+n ((n),(n) ))) =((23)/(11)) n=?

Question Number 208263 Answers: 1 Comments: 0

Question Number 208238 Answers: 1 Comments: 0

Show that (π/4) < ∫_0 ^1 (√(1−x^4 ))dx using x = sint show that ∫_0 ^1 (√(1−x^4 ))dx<((2(√2))/3) using (∫_0 ^1 f(x)g(x)dx)^2 <∫_0 ^1 (f(x))^2 dx∫_0 ^1 (g(x))^2 dx

$$\mathrm{S}{how}\:{that} \\ $$$$\frac{\pi}{\mathrm{4}}\:<\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:{using}\:{x}\:=\:{sint} \\ $$$${show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}<\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${using}\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){g}\left({x}\right){dx}\right)^{\mathrm{2}} <\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\int_{\mathrm{0}} ^{\mathrm{1}} \left({g}\left({x}\right)\right)^{\mathrm{2}} {dx} \\ $$

Question Number 207937 Answers: 0 Comments: 0

An ordered data consists of 6 even numbers and 4 odd numbers. The average of the odd numbers is 2022. The 3rd, 5th, 6th and 8 th numbers are odd. The data range is 24 , and the interquartile range is 14. The largest possible average of the ten numbers is ___

$$\:\:\mathrm{An}\:\mathrm{ordered}\:\mathrm{data}\:\mathrm{consists}\:\mathrm{of}\: \\ $$$$\:\mathrm{6}\:\mathrm{even}\:\mathrm{numbers}\:\mathrm{and}\:\mathrm{4}\:\mathrm{odd}\:\mathrm{numbers}. \\ $$$$\:\mathrm{The}\:\mathrm{average}\:\mathrm{of}\:\mathrm{the}\:\mathrm{odd}\:\mathrm{numbers}\: \\ $$$$\:\mathrm{is}\:\mathrm{2022}.\:\mathrm{The}\:\mathrm{3rd},\:\mathrm{5th},\:\mathrm{6th}\:\mathrm{and} \\ $$$$\:\mathrm{8}\:\mathrm{th}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{odd}.\: \\ $$$$\:\mathrm{The}\:\mathrm{data}\:\mathrm{range}\:\mathrm{is}\:\mathrm{24}\:,\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\:\mathrm{interquartile}\:\mathrm{range}\:\mathrm{is}\:\mathrm{14}.\: \\ $$$$\:\mathrm{The}\:\mathrm{largest}\:\mathrm{possible}\:\mathrm{average} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{ten}\:\mathrm{numbers}\:\mathrm{is}\:\_\_\_\: \\ $$

Question Number 207179 Answers: 2 Comments: 0