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Find-number-of-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-




Question Number 49482 by rahul 19 last updated on 07/Dec/18
Find number of 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 ?
$${Find}\:{number}\:{of}\:\mathrm{4}−{letter}\:{words}\:{which} \\ $$$${can}\:{be}\:{formed}\:{using}\:{the}\:{letters}\:{of}\: \\ $$$${the}\:{word}\:'{ALLAHABAD}'\:{such}\:{that}: \\ $$$${NO}\:{repetition}\:{of}\:{letter}\:{and}\:{word}\:{should} \\ $$$${start}\:“{either}''\:{from}\:{H}\:“{or}''\:{ends}\:{with}\:{D}\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
what is the answer...
$${what}\:{is}\:{the}\:{answer}… \\ $$
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 ??
$${I}'{m}\:\:{confused}\:{in}\:{the}\:{language}\:{of}\:{the} \\ $$$${problem}…. \\ $$$${If}\:{a}\:{word}\:{starts}\:{from}\:{H}\:{and}\:{ends}\:{with} \\ $$$${D}.\:{Is}\:{this}\:{case}\:{acceptable}\:?? \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
there are 5 letters A,L,H,B,D  (repetion not allowed)  we have to make 4 letter words in which one   letter is fixed. So we actually have to make  3 letter words with remaining letters.  which is^4 P_3 =((4!)/((4−3)!))=4!=24 words.
$${there}\:{are}\:\mathrm{5}\:{letters}\:{A},{L},{H},{B},{D}\:\:\left({repetion}\:{not}\:{allowed}\right) \\ $$$${we}\:{have}\:{to}\:{make}\:\mathrm{4}\:{letter}\:{words}\:{in}\:{which}\:{one}\: \\ $$$${letter}\:{is}\:{fixed}.\:{So}\:{we}\:{actually}\:{have}\:{to}\:{make} \\ $$$$\mathrm{3}\:{letter}\:{words}\:{with}\:{remaining}\:{letters}. \\ $$$${which}\:{is}\:^{\mathrm{4}} {P}_{\mathrm{3}} =\frac{\mathrm{4}!}{\left(\mathrm{4}−\mathrm{3}\right)!}=\mathrm{4}!=\mathrm{24}\:{words}. \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
if the 4 letter starts with H and ends with D  then in the 4 letter word 2 places are fixed. We  have to arrage remaining 2 places with  remaining 3 letters.  no. of words=^3 P_2 =6  which are  HALD,HABD,HLAD,HLBD,HBAD,HBLD
$${if}\:{the}\:\mathrm{4}\:{letter}\:{starts}\:{with}\:{H}\:{and}\:{ends}\:{with}\:{D} \\ $$$${then}\:{in}\:{the}\:\mathrm{4}\:{letter}\:{word}\:\mathrm{2}\:{places}\:{are}\:{fixed}.\:{We} \\ $$$${have}\:{to}\:{arrage}\:{remaining}\:\mathrm{2}\:{places}\:{with} \\ $$$${remaining}\:\mathrm{3}\:{letters}. \\ $$$${no}.\:{of}\:{words}=\:^{\mathrm{3}} {P}_{\mathrm{2}} =\mathrm{6} \\ $$$${which}\:{are} \\ $$$${HALD},{HABD},{HLAD},{HLBD},{HBAD},{HBLD} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
yes language is confusing and I solved  it as two cases 1)starts from H, 2)ends with D
$${yes}\:{language}\:{is}\:{confusing}\:{and}\:{I}\:{solved} \\ $$$$\left.{i}\left.{t}\:{as}\:{two}\:{cases}\:\mathrm{1}\right){starts}\:{from}\:{H},\:\mathrm{2}\right){ends}\:{with}\:{D} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
24 i think
$$\mathrm{24}\:{i}\:{think} \\ $$
Commented by rahul 19 last updated on 07/Dec/18
Correct answer is 42 ( including the  cases when letter start from H & ends  with D).
$${Correct}\:{answer}\:{is}\:\mathrm{42}\:\left(\:{including}\:{the}\right. \\ $$$${cases}\:{when}\:{letter}\:{start}\:{from}\:{H}\:\&\:{ends} \\ $$$$\left.{with}\:{D}\right). \\ $$
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...
$${i}\:{am}\:{asking}\:{the}\:{final}\:{results}\:{not}\:{the}\:{methods}… \\ $$
Commented by mr W last updated on 07/Dec/18
the answer is 2×4!−3!=42  we have 5 different letters.  for words starting with H there are  4×3×2=4! possobilities  for words ending with D there are  also 4! possobilities, but some of them  start with H which are already counted.  number of words starting with H and ending with  D is 3×2=3!  ⇒ solution is 2×4!−3!=42
$${the}\:{answer}\:{is}\:\mathrm{2}×\mathrm{4}!−\mathrm{3}!=\mathrm{42} \\ $$$${we}\:{have}\:\mathrm{5}\:{different}\:{letters}. \\ $$$${for}\:{words}\:{starting}\:{with}\:{H}\:{there}\:{are} \\ $$$$\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{4}!\:{possobilities} \\ $$$${for}\:{words}\:{ending}\:{with}\:{D}\:{there}\:{are} \\ $$$${also}\:\mathrm{4}!\:{possobilities},\:{but}\:{some}\:{of}\:{them} \\ $$$${start}\:{with}\:{H}\:{which}\:{are}\:{already}\:{counted}. \\ $$$${number}\:{of}\:{words}\:{starting}\:{with}\:{H}\:{and}\:{ending}\:{with} \\ $$$${D}\:{is}\:\mathrm{3}×\mathrm{2}=\mathrm{3}! \\ $$$$\Rightarrow\:{solution}\:{is}\:\mathrm{2}×\mathrm{4}!−\mathrm{3}!=\mathrm{42} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$
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.
$${yes}.\:{i}\:{have}\:{counted}\:{also}\:{words}\:{starting} \\ $$$${with}\:{H}\:{and}\:{ending}\:{with}\:{D}.\:{i}\:{think} \\ $$$${the}\:{question}\:{means}\:{words}\:{starting} \\ $$$${with}\:{H},\:{no}\:{matter}\:{if}\:{they}\:{end}\:{with}\:{D} \\ $$$${or}\:{not},\:{or}\:{words}\:{ending}\:{with}\:{D},\:{no} \\ $$$${matter}\:{if}\:{they}\:{start}\:{with}\:{H}\:{or}\:{not}. \\ $$
Commented by mr W last updated on 07/Dec/18
i think the language is quite clear:  either condition 1 or condition 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 1 or condition 2, but  not condition 1 and  condition 2 together
$${i}\:{think}\:{the}\:{language}\:{is}\:{quite}\:{clear}: \\ $$$${either}\:{condition}\:\mathrm{1}\:{or}\:{condition}\:\mathrm{2} \\ $$$${means}\:{one}\:{of}\:{the}\:{conditions} \\ $$$${must}\:{be}\:{fulfilled}.\:{it}\:{doesn}'{t}\:{reject}\: \\ $$$${that}\:{both}\:{conditions}\:{are}\:{fulfilled}.\:{if} \\ $$$${it}\:{is}\:{not}\:{allowed}\:{that}\:{both}\:{conditions} \\ $$$${are}\:{fulfilled},\:{then}\:{the}\:{question}\:{has} \\ $$$${had}\:{to}\:{be}: \\ $$$${either}\:{condition}\:\mathrm{1}\:{or}\:{condition}\:\mathrm{2},\:{but} \\ $$$${not}\:{condition}\:\mathrm{1}\:{and}\:\:{condition}\:\mathrm{2}\:{together} \\ $$
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=4  B=1   D=1  H=1  L=2  1)words start with H  H_1  −_2  −_3  −_4   selection of alphabet   one A from 4 A  one B from 1 B  one D from 1 D  one L from 2 L   so vacant place three(03) but avail able alphabet  (A,B,D,L) so permutation=4×3×2=24  (start by H  end by D)+(start by H end by others) =24  2)words end with D  −_1  −_2  −_3  D  one A from 4A  one B from 1B  one H from 1H  one L from 2L  sl vacant place three(03) available alphabets  (A, B,H,L) so permutation=4×3×2=24  (end by D start byH+end by D start by other)=24  3)words start by H and end by D  H_1   −_2   −_3  D_4    one A from 4A  one B from 1B  one L from 2L  vacant place two(02) available alphbets  (A,B,L) permutation 3×2=6  (start H and end by D)=6    P or Q=P∪Q=P+Q−P∩Q  P and Q=P∩Q  so results is 24+24−6=42
$${A}=\mathrm{4}\:\:{B}=\mathrm{1}\:\:\:{D}=\mathrm{1}\:\:{H}=\mathrm{1}\:\:{L}=\mathrm{2} \\ $$$$\left.\mathrm{1}\right){words}\:{start}\:{with}\:{H} \\ $$$$\underset{\mathrm{1}} {{H}}\:\underset{\mathrm{2}} {−}\:\underset{\mathrm{3}} {−}\:\underset{\mathrm{4}} {−} \\ $$$${selection}\:{of}\:{alphabet}\: \\ $$$${one}\:{A}\:{from}\:\mathrm{4}\:{A} \\ $$$${one}\:{B}\:{from}\:\mathrm{1}\:{B} \\ $$$${one}\:{D}\:{from}\:\mathrm{1}\:{D} \\ $$$${one}\:{L}\:{from}\:\mathrm{2}\:{L}\: \\ $$$${so}\:{vacant}\:{place}\:{three}\left(\mathrm{03}\right)\:{but}\:{avail}\:{able}\:{alphabet} \\ $$$$\left({A},{B},{D},{L}\right)\:{so}\:{permutation}=\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{24} \\ $$$$\left({start}\:{by}\:{H}\:\:{end}\:{by}\:{D}\right)+\left({start}\:{by}\:{H}\:{end}\:{by}\:{others}\right)\:=\mathrm{24} \\ $$$$\left.\mathrm{2}\right){words}\:{end}\:{with}\:{D} \\ $$$$\underset{\mathrm{1}} {−}\:\underset{\mathrm{2}} {−}\:\underset{\mathrm{3}} {−}\:{D} \\ $$$${one}\:{A}\:{from}\:\mathrm{4}{A} \\ $$$${one}\:{B}\:{from}\:\mathrm{1}{B} \\ $$$${one}\:{H}\:{from}\:\mathrm{1}{H} \\ $$$${one}\:{L}\:{from}\:\mathrm{2}{L} \\ $$$${sl}\:{vacant}\:{place}\:{three}\left(\mathrm{03}\right)\:{available}\:{alphabets} \\ $$$$\left({A},\:{B},{H},{L}\right)\:{so}\:{permutation}=\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{24} \\ $$$$\left({end}\:{by}\:{D}\:{start}\:{byH}+{end}\:{by}\:{D}\:{start}\:{by}\:{other}\right)=\mathrm{24} \\ $$$$\left.\mathrm{3}\right){words}\:{start}\:{by}\:{H}\:{and}\:{end}\:{by}\:{D} \\ $$$$\underset{\mathrm{1}} {{H}}\:\:\underset{\mathrm{2}} {−}\:\:\underset{\mathrm{3}} {−}\:\underset{\mathrm{4}} {{D}}\: \\ $$$${one}\:{A}\:{from}\:\mathrm{4}{A} \\ $$$${one}\:{B}\:{from}\:\mathrm{1}{B} \\ $$$${one}\:{L}\:{from}\:\mathrm{2}{L} \\ $$$${vacant}\:{place}\:{two}\left(\mathrm{02}\right)\:{available}\:{alphbets} \\ $$$$\left({A},{B},{L}\right)\:{permutation}\:\mathrm{3}×\mathrm{2}=\mathrm{6} \\ $$$$\left({start}\:{H}\:{and}\:{end}\:{by}\:{D}\right)=\mathrm{6} \\ $$$$ \\ $$$${P}\:{or}\:{Q}={P}\cup{Q}={P}+{Q}−{P}\cap{Q} \\ $$$${P}\:{and}\:{Q}={P}\cap{Q} \\ $$$${so}\:{results}\:{is}\:\mathrm{24}+\mathrm{24}−\mathrm{6}=\mathrm{42} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
thanks...
$${thanks}… \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
loved how you used sets
$${loved}\:{how}\:{you}\:{used}\:{sets} \\ $$
Commented by rahul 19 last updated on 07/Dec/18
thank you sir! Finally, By Vienn diagram or , and becomes more clear!

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