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Question-118555




Question Number 118555 by bramlexs22 last updated on 18/Oct/20
Answered by benjo_mathlover last updated on 18/Oct/20
Start with the 4 non−red slots, then we  have  ((5),(2) ) ways in which we can assigned of  inserting the red slots to get a sequence of  8 categorised slots where the red slots   are all apart . So the number of ways is    ((5),(2) )×4!×4! = 5760 ways
$${Start}\:{with}\:{the}\:\mathrm{4}\:{non}−{red}\:{slots},\:{then}\:{we} \\ $$$${have}\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\:{ways}\:{in}\:{which}\:{we}\:{can}\:{assigned}\:{of} \\ $$$${inserting}\:{the}\:{red}\:{slots}\:{to}\:{get}\:{a}\:{sequence}\:{of} \\ $$$$\mathrm{8}\:{categorised}\:{slots}\:{where}\:{the}\:{red}\:{slots}\: \\ $$$${are}\:{all}\:{apart}\:.\:{So}\:{the}\:{number}\:{of}\:{ways}\:{is}\: \\ $$$$\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}×\mathrm{4}!×\mathrm{4}!\:=\:\mathrm{5760}\:{ways} \\ $$
Answered by mr W last updated on 18/Oct/20
⊠■■■■■■■⊠  ■ place for one red book  ■ place for one or more other books  ⊠ place for zero or more other books    (1+x+x^2 +...)^2 (x+x^2 +...)^3   =(x^3 /((1−x)^5 ))=x^3 Σ_(k=0) ^∞ C_4 ^(k+4) x^k   coef. of x^4  is C_4 ^5 =C_1 ^5 =5  ⇒number of ways =5×4!×4!=2880
$$\boxtimes\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\boxtimes \\ $$$$\blacksquare\:{place}\:{for}\:{one}\:{red}\:{book} \\ $$$$\blacksquare\:{place}\:{for}\:{one}\:{or}\:{more}\:{other}\:{books} \\ $$$$\boxtimes\:{place}\:{for}\:{zero}\:{or}\:{more}\:{other}\:{books} \\ $$$$ \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} \left({x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \\ $$$$=\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }={x}^{\mathrm{3}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{4}} \:{is}\:{C}_{\mathrm{4}} ^{\mathrm{5}} ={C}_{\mathrm{1}} ^{\mathrm{5}} =\mathrm{5} \\ $$$$\Rightarrow{number}\:{of}\:{ways}\:=\mathrm{5}×\mathrm{4}!×\mathrm{4}!=\mathrm{2880} \\ $$
Commented by prakash jain last updated on 18/Oct/20
This can result in a blank space.   ^5 C_4   is choosing 4 out of 5 blanks  □■□■□■□■□  and it can result in choosing Xs below  X■X■□■X■X  Generating Function Approach  □■□■■■■■■  The arrangement above will also  result in same function.  □■■■■■■■□  Since 3 books slot cannot be chosen  we only need to find coefficient on x^1   from 2 white spaces.  (1−x)^(−2)   =2
$$\mathrm{This}\:\mathrm{can}\:\mathrm{result}\:\mathrm{in}\:\mathrm{a}\:\mathrm{blank}\:\mathrm{space}.\: \\ $$$$\:^{\mathrm{5}} {C}_{\mathrm{4}} \\ $$$$\mathrm{is}\:\mathrm{choosing}\:\mathrm{4}\:\mathrm{out}\:\mathrm{of}\:\mathrm{5}\:\mathrm{blanks} \\ $$$$\Box\blacksquare\Box\blacksquare\Box\blacksquare\Box\blacksquare\Box \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{can}\:\mathrm{result}\:\mathrm{in}\:\mathrm{choosing}\:\mathrm{X}{s}\:\mathrm{below} \\ $$$$\mathrm{X}\blacksquare\mathrm{X}\blacksquare\Box\blacksquare\mathrm{X}\blacksquare\mathrm{X} \\ $$$$\boldsymbol{\mathrm{Generating}}\:\boldsymbol{\mathrm{Function}}\:\boldsymbol{\mathrm{Approach}} \\ $$$$\Box\blacksquare\Box\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \\ $$$$\mathrm{The}\:\mathrm{arrangement}\:\mathrm{above}\:\mathrm{will}\:\mathrm{also} \\ $$$$\mathrm{result}\:\mathrm{in}\:\mathrm{same}\:\mathrm{function}. \\ $$$$\Box\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\Box \\ $$$$\mathrm{Since}\:\mathrm{3}\:\mathrm{books}\:\mathrm{slot}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{chosen} \\ $$$$\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{coefficient}\:\mathrm{on}\:{x}^{\mathrm{1}} \\ $$$$\mathrm{from}\:\mathrm{2}\:\mathrm{white}\:\mathrm{spaces}. \\ $$$$\left(\mathrm{1}−{x}\right)^{−\mathrm{2}} \:\:=\mathrm{2} \\ $$
Commented by mr W last updated on 18/Oct/20
if the question were: 4 red books and  8 other books, then the answer is  C_4 ^9 ×4!×8!.  in such a case the generating function  method is the best approach i think.
$${if}\:{the}\:{question}\:{were}:\:\mathrm{4}\:{red}\:{books}\:{and} \\ $$$$\mathrm{8}\:{other}\:{books},\:{then}\:{the}\:{answer}\:{is} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{9}} ×\mathrm{4}!×\mathrm{8}!. \\ $$$${in}\:{such}\:{a}\:{case}\:{the}\:{generating}\:{function} \\ $$$${method}\:{is}\:{the}\:{best}\:{approach}\:{i}\:{think}. \\ $$
Commented by prakash jain last updated on 18/Oct/20
ok. I have deleted wrong answer to  avoid confusion to other people  who read. Just pasting relevant  info here.  ■■■■■■■■  ■■■■■■■■  ■■■■■■■■  ■■■■■■■■  ■■■■■■■■  The above are 5 different ways  red books can be arranged.
$$\mathrm{ok}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{deleted}\:\mathrm{wrong}\:\mathrm{answer}\:\mathrm{to} \\ $$$$\mathrm{avoid}\:\mathrm{confusion}\:\mathrm{to}\:\mathrm{other}\:\mathrm{people} \\ $$$$\mathrm{who}\:\mathrm{read}.\:\mathrm{Just}\:\mathrm{pasting}\:\mathrm{relevant} \\ $$$$\mathrm{info}\:\mathrm{here}. \\ $$$$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \\ $$$$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \\ $$$$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \\ $$$$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \\ $$$$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{are}\:\mathrm{5}\:\mathrm{different}\:\mathrm{ways} \\ $$$$\mathrm{red}\:\mathrm{books}\:\mathrm{can}\:\mathrm{be}\:\mathrm{arranged}. \\ $$
Commented by mr W last updated on 18/Oct/20
absolutely right sir! thanks alot!
$${absolutely}\:{right}\:{sir}!\:{thanks}\:{alot}! \\ $$
Commented by bramlexs22 last updated on 19/Oct/20
but in my books the answer is 5760 sir.
$${but}\:{in}\:{my}\:{books}\:{the}\:{answer}\:{is}\:\mathrm{5760}\:{sir}. \\ $$$$ \\ $$

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