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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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