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Permutation and CombinationQuestion and Answers: Page 7

Question Number 169159 Answers: 1 Comments: 1

The remainder of 13^(13) when divided by 99 is

$${The}\:{remainder}\:{of} \\ $$$$\mathrm{13}^{\mathrm{13}} {when}\: \\ $$$${divided}\:{by}\:\mathrm{99}\:{is} \\ $$

Question Number 168291 Answers: 2 Comments: 0

∫t^7 sin(t^7 )dt

$$\int{t}^{\mathrm{7}} \mathrm{sin}\left({t}^{\mathrm{7}} \right){dt} \\ $$

Question Number 167979 Answers: 1 Comments: 0

3 men and 4 women are to sit on a table. Calculate the number of possible sitting arrangements if (a) they sit in a row such that the men must not sit next to each other. (b) they sit in circular pattern and the clockwise and anticlockwise orders are considered the same.

$$\mathrm{3}\:\mathrm{men}\:\mathrm{and}\:\mathrm{4}\:\mathrm{women}\:\mathrm{are}\:\mathrm{to}\:\mathrm{sit}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{table}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{possible}\:\mathrm{sitting}\:\mathrm{arrangements}\:\mathrm{if} \\ $$$$\:\left({a}\right)\:\mathrm{they}\:\mathrm{sit}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{men}\:\mathrm{must}\:\mathrm{not}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{other}. \\ $$$$\:\left({b}\right)\:\mathrm{they}\:\mathrm{sit}\:\mathrm{in}\:\mathrm{circular}\:\mathrm{pattern}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{the}\:\mathrm{clockwise}\:\mathrm{and}\:\mathrm{anticlockwise} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{orders}\:\mathrm{are}\:\mathrm{considered}\:\mathrm{the}\:\mathrm{same}. \\ $$

Question Number 167765 Answers: 0 Comments: 0

show that two permutations are conjugate if their matrices are similar

$$ \\ $$show that two permutations are conjugate if their matrices are similar

Question Number 166787 Answers: 0 Comments: 3

How many permutations of the letters of the word EINSTEIN are possible if the EIN groups must not be next to eachother?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{permutations}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}\:\mathrm{EINSTEIN}\:\mathrm{are} \\ $$$$\mathrm{possible}\:\mathrm{if}\:\mathrm{the}\:\mathrm{EIN}\:\mathrm{groups}\:\mathrm{must}\:\mathrm{not} \\ $$$$\mathrm{be}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}? \\ $$

Question Number 166770 Answers: 4 Comments: 6

In how many ways can you go up a staircase with 20 steps if you take one, two or three steps at a time?

$${In}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{go}\:{up}\:{a}\: \\ $$$${staircase}\:{with}\:\mathrm{20}\:{steps}\:{if}\:{you}\:{take}\: \\ $$$${one},\:{two}\:{or}\:{three}\:{steps}\:{at}\:{a}\:{time}? \\ $$

Question Number 166569 Answers: 0 Comments: 0

Question Number 165385 Answers: 1 Comments: 1

Number of ways..n differrnt things be distributed in r identical boxes so as 1)empty box is allowed 2)empty box not allowed Number of ways...n identical things be distributed to r identical boxes so as 1)empty box allowed 2)empty box not allowed

$${Number}\:{of}\:\:{ways}..{n}\: \\ $$$${differrnt}\:\:{things}\:{be}\:{distributed} \\ $$$${in}\:{r}\:{identical}\:{boxes}\:{so}\:{as} \\ $$$$\left.\mathrm{1}\right){empty}\:{box}\:{is}\:{allowed} \\ $$$$\left.\mathrm{2}\right){empty}\:{box}\:{not}\:{allowed} \\ $$$${Number}\:{of}\:{ways}...{n}\:{identical} \\ $$$${things}\:{be}\:{distributed}\:{to}\:{r} \\ $$$${identical}\:{boxes}\:{so}\:{as} \\ $$$$\left.\mathrm{1}\right){empty}\:{box}\:{allowed} \\ $$$$\left.\mathrm{2}\right){empty}\:{box}\:{not}\:{allowed} \\ $$$$ \\ $$

Question Number 165376 Answers: 1 Comments: 0

Verify wether f is invertible f (x) = (1+2x)^3

$$\mathrm{Verify}\:\mathrm{wether}\:{f}\:\mathrm{is}\:\mathrm{invertible}\: \\ $$$${f}\:\left({x}\right)\:=\:\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$

Question Number 165361 Answers: 1 Comments: 2

Obtain a general formula for the sequence (2/3),(4/5),(8/9),((16)/(17)),((32)/(33)),... assuming the sequence continues in that pattern.

$$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{general}\:\mathrm{formula}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{sequence} \\ $$$$\:\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{9}},\frac{\mathrm{16}}{\mathrm{17}},\frac{\mathrm{32}}{\mathrm{33}},... \\ $$$$\mathrm{assuming}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{continues}\:\mathrm{in}\:\mathrm{that} \\ $$$$\mathrm{pattern}. \\ $$

Question Number 164996 Answers: 1 Comments: 0

Question Number 164588 Answers: 0 Comments: 0

soit K un corps; pour toute permutation σ de S_n , on note P(σ) sa matrice dans la base canonique de K^n . montrer que deux permutations σ_1 et σ_2 sont conjugues dans S_n si et seulement si P(σ_1 ) et P(σ_2 ) sont semblables.

$${soit}\:{K}\:{un}\:{corps};\:{pour}\:{toute}\:{permutation} \\ $$$$\sigma\:{de}\:{S}_{{n}} ,\:{on}\:{note}\:{P}\left(\sigma\right)\:{sa}\:{matrice}\:{dans}\:{la}\:{base} \\ $$$${canonique}\:{de}\:{K}^{{n}} . \\ $$$${montrer}\:{que}\:{deux}\:{permutations}\:\sigma_{\mathrm{1}} \:{et}\:\sigma_{\mathrm{2}} \:{sont} \\ $$$${conjugues}\:{dans}\:{S}_{{n}} \:{si}\:{et}\:{seulement}\:{si}\: \\ $$$${P}\left(\sigma_{\mathrm{1}} \right)\:{et}\:{P}\left(\sigma_{\mathrm{2}} \right)\:{sont}\:{semblables}. \\ $$

Question Number 164544 Answers: 2 Comments: 0

prove that Σ_(k=1) ^n ((n),(k) )^2 =(((2n)!)/((n!)^2 ))−1

$${prove}\:{that}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }−\mathrm{1} \\ $$

Question Number 163443 Answers: 0 Comments: 7

p ≤ n find (A_n ^p /A_(n−1) ^p ).

$$\mathrm{p}\:\leqslant\:\mathrm{n}\: \\ $$$$\mathrm{find}\:\:\frac{\mathrm{A}_{\mathrm{n}} ^{\mathrm{p}} }{\mathrm{A}_{\mathrm{n}−\mathrm{1}} ^{\mathrm{p}} }. \\ $$

Question Number 163313 Answers: 1 Comments: 0

Find the non negative integer solutions of 2x+3y+5z=60

$${Find}\:{the}\:{non}\:{negative}\:{integer} \\ $$$${solutions}\:{of}\:\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{5}{z}=\mathrm{60} \\ $$

Question Number 163283 Answers: 1 Comments: 1

hi ! We store 5 objects in three discernible drawers. Suppose that the different ways of carrying out these arrangements are equally probable, calculate the probability that one of the 3 drawers contains at least 3 objects.

$$\mathrm{hi}\:! \\ $$We store 5 objects in three discernible drawers. Suppose that the different ways of carrying out these arrangements are equally probable, calculate the probability that one of the 3 drawers contains at least 3 objects.

Question Number 163263 Answers: 1 Comments: 0

Question Number 160980 Answers: 2 Comments: 0

How many ways can 50 people be divided into 3 groups, so that each group contains members equal to a prime number?

$$\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{50}\:\mathrm{people} \\ $$$$\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{3}\:\mathrm{groups},\:\mathrm{so} \\ $$$$\:\mathrm{that}\:\mathrm{each}\:\mathrm{group}\:\mathrm{contains}\:\mathrm{members} \\ $$$$\:\mathrm{equal}\:\mathrm{to}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}? \\ $$

Question Number 160829 Answers: 0 Comments: 1

Question Number 160665 Answers: 0 Comments: 2

In how many ways can you divide 20 students into 5 groups with 4 students in each group such that any two students don′t meet each other in more than one group.

$${In}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{divide}\:\mathrm{20} \\ $$$${students}\:{into}\:\mathrm{5}\:{groups}\:{with}\:\mathrm{4}\:{students} \\ $$$${in}\:{each}\:{group}\:{such}\:{that}\:{any}\:{two} \\ $$$${students}\:{don}'{t}\:{meet}\:{each}\:{other}\:{in} \\ $$$${more}\:{than}\:{one}\:{group}. \\ $$

Question Number 160389 Answers: 0 Comments: 2

Question Number 159715 Answers: 0 Comments: 1

Question Number 159259 Answers: 1 Comments: 0

montre que Σ_(k=0) ^n C_(2n) ^(2k) =2^(2n−1)

$$\boldsymbol{{montre}}\:\boldsymbol{{que}}\: \\ $$$$\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{C}}_{\mathrm{2}\boldsymbol{{n}}} ^{\mathrm{2}\boldsymbol{{k}}} =\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}} \\ $$

Question Number 158955 Answers: 1 Comments: 0

Question Number 158529 Answers: 1 Comments: 0

In how many ways can 30 students be distributed to 10 schools, if 1. each school should get at least one student. 2. no restriction

$${In}\:{how}\:{many}\:{ways}\:{can}\:\mathrm{30}\:{students} \\ $$$${be}\:{distributed}\:{to}\:\mathrm{10}\:{schools},\:{if} \\ $$$$\mathrm{1}.\:{each}\:{school}\:{should}\:{get}\:{at}\:{least}\:{one} \\ $$$$\:\:\:\:\:{student}. \\ $$$$\mathrm{2}.\:{no}\:{restriction} \\ $$

Question Number 157980 Answers: 2 Comments: 0

Calculate this problem below a). ((7!)/((5−1)!)) =...... b). ((5!×3!)/(4!)) =...... c). 6! 4! =... d). ((7!)/(3!))×((2!)/(5!)) =......

$$\mathrm{Calculate}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{below} \\ $$$$\left.\:\:\:\:\:\:\:\:\mathrm{a}\right).\:\:\frac{\mathrm{7}!}{\left(\mathrm{5}−\mathrm{1}\right)!}\:=...... \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\mathrm{b}\right).\:\:\frac{\mathrm{5}!×\mathrm{3}!}{\mathrm{4}!}\:=...... \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}\right).\:\:\mathrm{6}!\:\mathrm{4}!\:=...\:\:\:\:\:\:\:\: \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\mathrm{d}\right).\:\:\frac{\mathrm{7}!}{\mathrm{3}!}×\frac{\mathrm{2}!}{\mathrm{5}!}\:=...... \\ $$

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