Question Number 49482 by rahul 19 last updated on 07/Dec/18
$${Find}\:{number}\:{of}\:\mathrm{4}−{letter}\:{words}\:{which} \\ $$$${can}\:{be}\:{formed}\:{using}\:{the}\:{letters}\:{of}\: \\ $$$${the}\:{word}\:'{ALLAHABAD}'\:{such}\:{that}: \\ $$$${NO}\:{repetition}\:{of}\:{letter}\:{and}\:{word}\:{should} \\ $$$${start}\:“{either}''\:{from}\:{H}\:“{or}''\:{ends}\:{with}\:{D}\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${what}\:{is}\:{the}\:{answer}… \\ $$
Commented by rahul 19 last updated on 07/Dec/18
$${I}'{m}\:\:{confused}\:{in}\:{the}\:{language}\:{of}\:{the} \\ $$$${problem}…. \\ $$$${If}\:{a}\:{word}\:{starts}\:{from}\:{H}\:{and}\:{ends}\:{with} \\ $$$${D}.\:{Is}\:{this}\:{case}\:{acceptable}\:?? \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
$${there}\:{are}\:\mathrm{5}\:{letters}\:{A},{L},{H},{B},{D}\:\:\left({repetion}\:{not}\:{allowed}\right) \\ $$$${we}\:{have}\:{to}\:{make}\:\mathrm{4}\:{letter}\:{words}\:{in}\:{which}\:{one}\: \\ $$$${letter}\:{is}\:{fixed}.\:{So}\:{we}\:{actually}\:{have}\:{to}\:{make} \\ $$$$\mathrm{3}\:{letter}\:{words}\:{with}\:{remaining}\:{letters}. \\ $$$${which}\:{is}\:^{\mathrm{4}} {P}_{\mathrm{3}} =\frac{\mathrm{4}!}{\left(\mathrm{4}−\mathrm{3}\right)!}=\mathrm{4}!=\mathrm{24}\:{words}. \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
$${if}\:{the}\:\mathrm{4}\:{letter}\:{starts}\:{with}\:{H}\:{and}\:{ends}\:{with}\:{D} \\ $$$${then}\:{in}\:{the}\:\mathrm{4}\:{letter}\:{word}\:\mathrm{2}\:{places}\:{are}\:{fixed}.\:{We} \\ $$$${have}\:{to}\:{arrage}\:{remaining}\:\mathrm{2}\:{places}\:{with} \\ $$$${remaining}\:\mathrm{3}\:{letters}. \\ $$$${no}.\:{of}\:{words}=\:^{\mathrm{3}} {P}_{\mathrm{2}} =\mathrm{6} \\ $$$${which}\:{are} \\ $$$${HALD},{HABD},{HLAD},{HLBD},{HBAD},{HBLD} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
$${yes}\:{language}\:{is}\:{confusing}\:{and}\:{I}\:{solved} \\ $$$$\left.{i}\left.{t}\:{as}\:{two}\:{cases}\:\mathrm{1}\right){starts}\:{from}\:{H},\:\mathrm{2}\right){ends}\:{with}\:{D} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
$$\mathrm{24}\:{i}\:{think} \\ $$
Commented by rahul 19 last updated on 07/Dec/18
$${Correct}\:{answer}\:{is}\:\mathrm{42}\:\left(\:{including}\:{the}\right. \\ $$$${cases}\:{when}\:{letter}\:{start}\:{from}\:{H}\:\&\:{ends} \\ $$$$\left.{with}\:{D}\right). \\ $$
Commented by rahul 19 last updated on 07/Dec/18
thank you sir!
Sir, but why have you counted the cases in which word starts from H and ends with D ... That's my query.
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${i}\:{am}\:{asking}\:{the}\:{final}\:{results}\:{not}\:{the}\:{methods}… \\ $$
Commented by mr W last updated on 07/Dec/18
$${the}\:{answer}\:{is}\:\mathrm{2}×\mathrm{4}!−\mathrm{3}!=\mathrm{42} \\ $$$${we}\:{have}\:\mathrm{5}\:{different}\:{letters}. \\ $$$${for}\:{words}\:{starting}\:{with}\:{H}\:{there}\:{are} \\ $$$$\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{4}!\:{possobilities} \\ $$$${for}\:{words}\:{ending}\:{with}\:{D}\:{there}\:{are} \\ $$$${also}\:\mathrm{4}!\:{possobilities},\:{but}\:{some}\:{of}\:{them} \\ $$$${start}\:{with}\:{H}\:{which}\:{are}\:{already}\:{counted}. \\ $$$${number}\:{of}\:{words}\:{starting}\:{with}\:{H}\:{and}\:{ending}\:{with} \\ $$$${D}\:{is}\:\mathrm{3}×\mathrm{2}=\mathrm{3}! \\ $$$$\Rightarrow\:{solution}\:{is}\:\mathrm{2}×\mathrm{4}!−\mathrm{3}!=\mathrm{42} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by mr W last updated on 07/Dec/18
$${yes}.\:{i}\:{have}\:{counted}\:{also}\:{words}\:{starting} \\ $$$${with}\:{H}\:{and}\:{ending}\:{with}\:{D}.\:{i}\:{think} \\ $$$${the}\:{question}\:{means}\:{words}\:{starting} \\ $$$${with}\:{H},\:{no}\:{matter}\:{if}\:{they}\:{end}\:{with}\:{D} \\ $$$${or}\:{not},\:{or}\:{words}\:{ending}\:{with}\:{D},\:{no} \\ $$$${matter}\:{if}\:{they}\:{start}\:{with}\:{H}\:{or}\:{not}. \\ $$
Commented by mr W last updated on 07/Dec/18
$${i}\:{think}\:{the}\:{language}\:{is}\:{quite}\:{clear}: \\ $$$${either}\:{condition}\:\mathrm{1}\:{or}\:{condition}\:\mathrm{2} \\ $$$${means}\:{one}\:{of}\:{the}\:{conditions} \\ $$$${must}\:{be}\:{fulfilled}.\:{it}\:{doesn}'{t}\:{reject}\: \\ $$$${that}\:{both}\:{conditions}\:{are}\:{fulfilled}.\:{if} \\ $$$${it}\:{is}\:{not}\:{allowed}\:{that}\:{both}\:{conditions} \\ $$$${are}\:{fulfilled},\:{then}\:{the}\:{question}\:{has} \\ $$$${had}\:{to}\:{be}: \\ $$$${either}\:{condition}\:\mathrm{1}\:{or}\:{condition}\:\mathrm{2},\:{but} \\ $$$${not}\:{condition}\:\mathrm{1}\:{and}\:\:{condition}\:\mathrm{2}\:{together} \\ $$
Commented by rahul 19 last updated on 07/Dec/18
thank you so much sir!
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${A}=\mathrm{4}\:\:{B}=\mathrm{1}\:\:\:{D}=\mathrm{1}\:\:{H}=\mathrm{1}\:\:{L}=\mathrm{2} \\ $$$$\left.\mathrm{1}\right){words}\:{start}\:{with}\:{H} \\ $$$$\underset{\mathrm{1}} {{H}}\:\underset{\mathrm{2}} {−}\:\underset{\mathrm{3}} {−}\:\underset{\mathrm{4}} {−} \\ $$$${selection}\:{of}\:{alphabet}\: \\ $$$${one}\:{A}\:{from}\:\mathrm{4}\:{A} \\ $$$${one}\:{B}\:{from}\:\mathrm{1}\:{B} \\ $$$${one}\:{D}\:{from}\:\mathrm{1}\:{D} \\ $$$${one}\:{L}\:{from}\:\mathrm{2}\:{L}\: \\ $$$${so}\:{vacant}\:{place}\:{three}\left(\mathrm{03}\right)\:{but}\:{avail}\:{able}\:{alphabet} \\ $$$$\left({A},{B},{D},{L}\right)\:{so}\:{permutation}=\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{24} \\ $$$$\left({start}\:{by}\:{H}\:\:{end}\:{by}\:{D}\right)+\left({start}\:{by}\:{H}\:{end}\:{by}\:{others}\right)\:=\mathrm{24} \\ $$$$\left.\mathrm{2}\right){words}\:{end}\:{with}\:{D} \\ $$$$\underset{\mathrm{1}} {−}\:\underset{\mathrm{2}} {−}\:\underset{\mathrm{3}} {−}\:{D} \\ $$$${one}\:{A}\:{from}\:\mathrm{4}{A} \\ $$$${one}\:{B}\:{from}\:\mathrm{1}{B} \\ $$$${one}\:{H}\:{from}\:\mathrm{1}{H} \\ $$$${one}\:{L}\:{from}\:\mathrm{2}{L} \\ $$$${sl}\:{vacant}\:{place}\:{three}\left(\mathrm{03}\right)\:{available}\:{alphabets} \\ $$$$\left({A},\:{B},{H},{L}\right)\:{so}\:{permutation}=\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{24} \\ $$$$\left({end}\:{by}\:{D}\:{start}\:{byH}+{end}\:{by}\:{D}\:{start}\:{by}\:{other}\right)=\mathrm{24} \\ $$$$\left.\mathrm{3}\right){words}\:{start}\:{by}\:{H}\:{and}\:{end}\:{by}\:{D} \\ $$$$\underset{\mathrm{1}} {{H}}\:\:\underset{\mathrm{2}} {−}\:\:\underset{\mathrm{3}} {−}\:\underset{\mathrm{4}} {{D}}\: \\ $$$${one}\:{A}\:{from}\:\mathrm{4}{A} \\ $$$${one}\:{B}\:{from}\:\mathrm{1}{B} \\ $$$${one}\:{L}\:{from}\:\mathrm{2}{L} \\ $$$${vacant}\:{place}\:{two}\left(\mathrm{02}\right)\:{available}\:{alphbets} \\ $$$$\left({A},{B},{L}\right)\:{permutation}\:\mathrm{3}×\mathrm{2}=\mathrm{6} \\ $$$$\left({start}\:{H}\:{and}\:{end}\:{by}\:{D}\right)=\mathrm{6} \\ $$$$ \\ $$$${P}\:{or}\:{Q}={P}\cup{Q}={P}+{Q}−{P}\cap{Q} \\ $$$${P}\:{and}\:{Q}={P}\cap{Q} \\ $$$${so}\:{results}\:{is}\:\mathrm{24}+\mathrm{24}−\mathrm{6}=\mathrm{42} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${thanks}… \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
$${loved}\:{how}\:{you}\:{used}\:{sets} \\ $$
Commented by rahul 19 last updated on 07/Dec/18
thank you sir!
Finally, By Vienn diagram
or , and becomes more clear!