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Question-54286

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Question Number 54286 by Tinkutara last updated on 01/Feb/19
Commented by Tinkutara last updated on 01/Feb/19
Answers 4d 5b
Commented by mr W last updated on 05/Feb/19
4(b) 5(d) ???
$$\mathrm{4}\left({b}\right) \\ $$$$\mathrm{5}\left({d}\right) \\ $$$$??? \\ $$
Answered by mr W last updated on 02/Feb/19
Q4: number of numbers with one digit: C_1 ^9 number of numbers with two digits: C_2 ^9 number of numbers with 3 digits: C_3 ^9 ...... number of numbers with 9 digits: C_9 ^9 ⇒C_1 ^9 +C_2 ^9 +C_3 ^9 +...+C_9 ^9 =2^9 −1=511 ⇒(b)
$${Q}\mathrm{4}: \\ $$$${number}\:{of}\:{numbers}\:{with}\:{one}\:{digit}:\:{C}_{\mathrm{1}} ^{\mathrm{9}} \\ $$$${number}\:{of}\:{numbers}\:{with}\:{two}\:{digits}:\:{C}_{\mathrm{2}} ^{\mathrm{9}} \\ $$$${number}\:{of}\:{numbers}\:{with}\:\mathrm{3}\:{digits}:\:{C}_{\mathrm{3}} ^{\mathrm{9}} \\ $$$$…… \\ $$$${number}\:{of}\:{numbers}\:{with}\:\mathrm{9}\:{digits}:\:{C}_{\mathrm{9}} ^{\mathrm{9}} \\ $$$$ \\ $$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{9}} +{C}_{\mathrm{2}} ^{\mathrm{9}} +{C}_{\mathrm{3}} ^{\mathrm{9}} +…+{C}_{\mathrm{9}} ^{\mathrm{9}} =\mathrm{2}^{\mathrm{9}} −\mathrm{1}=\mathrm{511} \\ $$$$\Rightarrow\left({b}\right) \\ $$
Commented by Tinkutara last updated on 03/Feb/19
But in this are these numbers included like 2.1222333666 2.888999 2.446888 or any others with repetitons?
Commented by mr W last updated on 03/Feb/19
no repetition sir. the question requests non−zero digits in increasing order. 2.1245 is ok, but 2.12445 is not ok. if repetitions are allowed, there woude be infinite numbers, e.g. 2.12445, 2.124445, 2.1244445, etc.
$${no}\:{repetition}\:{sir}.\:{the}\:{question}\:{requests} \\ $$$${non}−{zero}\:{digits}\:{in}\:{increasing}\:{order}. \\ $$$$\mathrm{2}.\mathrm{1245}\:{is}\:{ok},\:{but}\:\mathrm{2}.\mathrm{12445}\:{is}\:{not}\:{ok}.\: \\ $$$${if}\:{repetitions}\:{are}\:{allowed},\:{there}\:{woude} \\ $$$${be}\:{infinite}\:{numbers},\:{e}.{g}.\:\mathrm{2}.\mathrm{12445}, \\ $$$$\mathrm{2}.\mathrm{124445},\:\mathrm{2}.\mathrm{1244445},\:{etc}. \\ $$
Answered by mr W last updated on 06/Feb/19
Q5: 3+(k−1)×3=93 ⇒k=31 ⇒option (d) explanation: such that the word MAT can not be formed, only one or two different letters may be selected. there are following ways to do this: all k coupons carry letter T: 1 way all k coupons carry letter A: 1 way all k coupons carry letter M: 1 way Σ: 3 ways k coupons with letters T and A: 1 coupon letter T, k−1 coupons letter A: 1 way 2 coupons letter T, k−2 coupons letter A: 1 way 3 coupons letter T, k−3 coupons letter A: 1 way ..... k−1 coupons letter T, 1 coupon letter A: 1 way Σ: (k−1) ways similarly k coupons with letters T and M: Σ: (k−1) ways k coupons with letters A and M: Σ: (k−1) ways totally there are 3+(k−1)×3 ways to select only one or two different letters. 3+(k−1)×3=3k=93 ⇒k=31
$${Q}\mathrm{5}: \\ $$$$\mathrm{3}+\left({k}−\mathrm{1}\right)×\mathrm{3}=\mathrm{93} \\ $$$$\Rightarrow{k}=\mathrm{31} \\ $$$$ \\ $$$$\Rightarrow{option}\:\left({d}\right) \\ $$$$ \\ $$$${explanation}: \\ $$$${such}\:{that}\:{the}\:{word}\:{MAT}\:{can}\:{not}\:{be}\:{formed}, \\ $$$${only}\:{one}\:{or}\:{two}\:{different} \\ $$$${letters}\:{may}\:{be}\:{selected}.\:{there}\:{are} \\ $$$${following}\:{ways}\:{to}\:{do}\:{this}: \\ $$$${all}\:{k}\:{coupons}\:{carry}\:{letter}\:{T}:\:\mathrm{1}\:{way} \\ $$$${all}\:{k}\:{coupons}\:{carry}\:{letter}\:{A}:\:\mathrm{1}\:{way} \\ $$$${all}\:{k}\:{coupons}\:{carry}\:{letter}\:{M}:\:\mathrm{1}\:{way} \\ $$$$\Sigma:\:\mathrm{3}\:{ways} \\ $$$$ \\ $$$${k}\:{coupons}\:{with}\:{letters}\:{T}\:{and}\:{A}: \\ $$$$\mathrm{1}\:{coupon}\:{letter}\:{T},\:{k}−\mathrm{1}\:{coupons}\:{letter}\:{A}:\:\mathrm{1}\:{way} \\ $$$$\mathrm{2}\:{coupons}\:{letter}\:{T},\:{k}−\mathrm{2}\:{coupons}\:{letter}\:{A}:\:\mathrm{1}\:{way} \\ $$$$\mathrm{3}\:{coupons}\:{letter}\:{T},\:{k}−\mathrm{3}\:{coupons}\:{letter}\:{A}:\:\mathrm{1}\:{way} \\ $$$$….. \\ $$$${k}−\mathrm{1}\:{coupons}\:{letter}\:{T},\:\mathrm{1}\:{coupon}\:{letter}\:{A}:\:\mathrm{1}\:{way} \\ $$$$\Sigma:\:\left({k}−\mathrm{1}\right)\:{ways} \\ $$$${similarly} \\ $$$${k}\:{coupons}\:{with}\:{letters}\:{T}\:{and}\:{M}: \\ $$$$\Sigma:\:\left({k}−\mathrm{1}\right)\:{ways} \\ $$$${k}\:{coupons}\:{with}\:{letters}\:{A}\:{and}\:{M}: \\ $$$$\Sigma:\:\left({k}−\mathrm{1}\right)\:{ways} \\ $$$$ \\ $$$${totally}\:{there}\:{are}\:\mathrm{3}+\left({k}−\mathrm{1}\right)×\mathrm{3}\:{ways} \\ $$$${to}\:{select}\:{only}\:{one}\:{or}\:{two}\:{different} \\ $$$${letters}. \\ $$$$\mathrm{3}+\left({k}−\mathrm{1}\right)×\mathrm{3}=\mathrm{3}{k}=\mathrm{93} \\ $$$$\Rightarrow{k}=\mathrm{31} \\ $$
Commented by Tinkutara last updated on 05/Feb/19
But sir no option is matching
Commented by mr W last updated on 06/Feb/19
answer is k=31. i.e. option d. proof that option b is wrong: if 5 coupons are selected, only following selections can′t spell the word MAT: AAAAA MMMMM TTTTT AMMMM AAMMM AAAMM AAAAM ATTTT AATTT AAATT AAAAT MTTTT MMTTT MMMTT MMMMT totally only 15 ways, not 93 ways. that is to say answer in book is wrong.
$${answer}\:{is}\:{k}=\mathrm{31}.\:{i}.{e}.\:{option}\:{d}. \\ $$$${proof}\:{that}\:{option}\:{b}\:{is}\:{wrong}: \\ $$$${if}\:\mathrm{5}\:{coupons}\:{are}\:{selected},\:{only}\:{following} \\ $$$${selections}\:{can}'{t}\:{spell}\:{the}\:{word}\:{MAT}: \\ $$$${AAAAA} \\ $$$${MMMMM} \\ $$$${TTTTT} \\ $$$${AMMMM} \\ $$$${AAMMM} \\ $$$${AAAMM} \\ $$$${AAAAM} \\ $$$${ATTTT} \\ $$$${AATTT} \\ $$$${AAATT} \\ $$$${AAAAT} \\ $$$${MTTTT} \\ $$$${MMTTT} \\ $$$${MMMTT} \\ $$$${MMMMT} \\ $$$${totally}\:{only}\:\mathrm{15}\:{ways},\:{not}\:\mathrm{93}\:{ways}. \\ $$$${that}\:{is}\:{to}\:{say}\:{answer}\:{in}\:{book}\:{is}\:{wrong}. \\ $$
Commented by Tinkutara last updated on 06/Feb/19
Thanks a lot Sir!

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