Menu Close

In-how-many-ways-can-the-word-KINECTIC-be-arranged-so-that-no-vowels-can-be-together-

Tinku Tara 0 Comments
Pin




Question Number 206939 by Tawa11 last updated on 01/May/24
In how many ways can the word KINECTIC be arranged so that no vowels can be together?
In how many ways can the word KINECTIC be
arranged so that no vowels can be together?
Commented by mr W last updated on 01/May/24
3600 ?
$$\mathrm{3600}\:? \\ $$
Commented by Tawa11 last updated on 01/May/24
I got same answer sir.
$$\mathrm{I}\:\mathrm{got}\:\mathrm{same}\:\mathrm{answer}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 01/May/24
But for KINETIC I got 2160 Textbook says 720
$$\mathrm{But}\:\mathrm{for}\:\mathrm{KINETIC} \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{2160} \\ $$$$\mathrm{Textbook}\:\mathrm{says}\:\:\mathrm{720} \\ $$
Commented by A5T last updated on 01/May/24
I guess the textbook is right about 720.
$${I}\:{guess}\:{the}\:{textbook}\:{is}\:{right}\:{about}\:\mathrm{720}. \\ $$
Commented by mr W last updated on 01/May/24
the question said KINECTIC, not KINETIC. if KINECTIC is meant, then answer is 3×20×((5!)/2)=3600. if KINETIC is meant, then answer is 3×10×4!=720.
$${the}\:{question}\:{said}\:{KINECTIC},\:\:{not} \\ $$$${KINETIC}. \\ $$$${if}\:{KINECTIC}\:{is}\:{meant},\:{then}\:{answer} \\ $$$${is}\:\mathrm{3}×\mathrm{20}×\frac{\mathrm{5}!}{\mathrm{2}}=\mathrm{3600}. \\ $$$${if}\:{KINETIC}\:{is}\:{meant},\:{then}\:{answer} \\ $$$${is}\:\mathrm{3}×\mathrm{10}×\mathrm{4}!=\mathrm{720}. \\ $$
Commented by BaliramKumar last updated on 01/May/24
solution_
$$\mathrm{solutio}\underset{} {\mathrm{n}} \\ $$
Commented by Tawa11 last updated on 01/May/24
Thanks sir. I understand now sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 01/May/24
KINECTIC the vowels (I, I, E) must be separated by at least one consonant. XVYVYVX V=one vowel X=place holder for zero or more consonants Y=place holder for one or more consonants to place the vowels there are ((3!)/(2!)) ways. to place the 5 consonants there are following possibilities: 0+1+1+3 0+1+2+2 0+1+3+1 0+1+4+0 0+2+1+2 0+2+2+1 0+2+3+0 0+3+1+1 0+3+2+0 0+4+1+0 1+1+1+2 1+1+2+1 1+1+3+0 1+2+1+1 1+2+2+0 1+3+1+0 2+1+1+1 2+1+2+0 2+2+1+0 3+1+1+0 that′s totally 20 possibilities. therefore there are ((3!)/(2!))×20×((5!)/(2!))=3600 ways similarly for KINETIC there are ((3!)/(2!))×10×4!=720 ways
$$\boldsymbol{{KINECTIC}} \\ $$$${the}\:{vowels}\:\left({I},\:{I},\:{E}\right)\:{must}\:{be}\:{separated}\: \\ $$$${by}\:{at}\:{least}\:{one}\:{consonant}. \\ $$$${XVYVYVX} \\ $$$${V}={one}\:{vowel} \\ $$$${X}={place}\:{holder}\:{for}\:{zero}\:{or}\:{more}\:{consonants} \\ $$$${Y}={place}\:{holder}\:{for}\:{one}\:{or}\:{more}\:{consonants} \\ $$$${to}\:{place}\:{the}\:{vowels}\:{there}\:{are}\:\frac{\mathrm{3}!}{\mathrm{2}!}\:{ways}.\: \\ $$$${to}\:{place}\:{the}\:\mathrm{5}\:{consonants}\:{there} \\ $$$${are}\:{following}\:{possibilities}: \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{1}+\mathrm{3} \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{2} \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{3}+\mathrm{1} \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{4}+\mathrm{0} \\ $$$$\mathrm{0}+\mathrm{2}+\mathrm{1}+\mathrm{2} \\ $$$$\mathrm{0}+\mathrm{2}+\mathrm{2}+\mathrm{1} \\ $$$$\mathrm{0}+\mathrm{2}+\mathrm{3}+\mathrm{0} \\ $$$$\mathrm{0}+\mathrm{3}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{0}+\mathrm{3}+\mathrm{2}+\mathrm{0} \\ $$$$\mathrm{0}+\mathrm{4}+\mathrm{1}+\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{2} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{2}+\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{3}+\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{2}+\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{1}+\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{2}+\mathrm{1}+\mathrm{2}+\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{2}+\mathrm{1}+\mathrm{0} \\ $$$$\mathrm{3}+\mathrm{1}+\mathrm{1}+\mathrm{0} \\ $$$${that}'{s}\:{totally}\:\mathrm{20}\:{possibilities}. \\ $$$${therefore}\:{there}\:{are} \\ $$$$\frac{\mathrm{3}!}{\mathrm{2}!}×\mathrm{20}×\frac{\mathrm{5}!}{\mathrm{2}!}=\mathrm{3600}\:{ways} \\ $$$$ \\ $$$${similarly}\:{for}\:\boldsymbol{{KINETIC}}\:{there}\:{are} \\ $$$$\frac{\mathrm{3}!}{\mathrm{2}!}×\mathrm{10}×\mathrm{4}!=\mathrm{720}\:{ways} \\ $$
Commented by mr W last updated on 02/May/24
Method II (generating function) for word KINECTIC XVYVYVX X⇒(1+x+x^2 +....) Y⇒(x+x^2 +....)=x(1+x+x^2 +...) (1+x+x^2 +....)^2 (x+x^2 +....)^2 =x^2 (1+x+x^2 +....)^4 =(x^2 /((1−x)^4 )) =x^2 Σ_(k=0) ^∞ C_3 ^(k+3) x^k coef. of x^5 term is C_3 ^(3+3) =20 ⇒number of ways: ((3!)/(2!))×20×((5!)/(2!))=3600 for word KINETIC coef. of x^4 term is C_3 ^(2+3) =10 ⇒number of ways: ((3!)/(2!))×10×4!=720
$${Method}\:{II}\:\left({generating}\:{function}\right) \\ $$$${for}\:{word}\:\boldsymbol{{KINECTIC}} \\ $$$${XVYVYVX} \\ $$$${X}\Rightarrow\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +….\right) \\ $$$${Y}\Rightarrow\left({x}+{x}^{\mathrm{2}} +….\right)={x}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +….\right)^{\mathrm{2}} \left({x}+{x}^{\mathrm{2}} +….\right)^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +….\right)^{\mathrm{4}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$={x}^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{term}\:{is}\:{C}_{\mathrm{3}} ^{\mathrm{3}+\mathrm{3}} =\mathrm{20} \\ $$$$\Rightarrow{number}\:{of}\:{ways}:\:\frac{\mathrm{3}!}{\mathrm{2}!}×\mathrm{20}×\frac{\mathrm{5}!}{\mathrm{2}!}=\mathrm{3600} \\ $$$$ \\ $$$${for}\:{word}\:\boldsymbol{{KINETIC}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{4}} \:{term}\:{is}\:{C}_{\mathrm{3}} ^{\mathrm{2}+\mathrm{3}} =\mathrm{10} \\ $$$$\Rightarrow{number}\:{of}\:{ways}:\:\frac{\mathrm{3}!}{\mathrm{2}!}×\mathrm{10}×\mathrm{4}!=\mathrm{720} \\ $$
Commented by Tawa11 last updated on 01/May/24
Great sir. I will study this.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{study}\:\mathrm{this}. \\ $$
Commented by mr W last updated on 02/May/24
generally if there are m (different) vowels and n (different) consonants (n+1≥m), then the answer is m!n!C_m ^(n+1) if there is repetition in vowels and or in consonants, then the answer is ((m!)/(m_1 !m_2 !...))×((n!)/(n_1 !n_2 !...))×C_m ^(n+1)
$${generally}\:{if}\:{there}\:{are}\:{m}\:\left({different}\right) \\ $$$${vowels}\:{and}\:{n}\:\left({different}\right)\:{consonants} \\ $$$$\left({n}+\mathrm{1}\geqslant{m}\right),\:{then}\:{the}\:{answer}\:{is} \\ $$$${m}!{n}!{C}_{{m}} ^{{n}+\mathrm{1}} \\ $$$${if}\:{there}\:{is}\:{repetition}\:{in}\:{vowels}\:{and} \\ $$$${or}\:{in}\:{consonants},\:{then}\:{the}\:{answer}\:{is} \\ $$$$\frac{{m}!}{{m}_{\mathrm{1}} !{m}_{\mathrm{2}} !…}×\frac{{n}!}{{n}_{\mathrm{1}} !{n}_{\mathrm{2}} !…}×{C}_{{m}} ^{{n}+\mathrm{1}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *