Menu Close

you-randomly-select-a-5-digit-number-what-s-the-probability-that-this-number-has-exactly-3-different-digits-

Tinku Tara 0 Comments
Pin




Question Number 188551 by mr W last updated on 03/Mar/23
you randomly select a 5 digit number. what′s the probability that this number has exactly 3 different digits?
$${you}\:{randomly}\:{select}\:{a}\:\mathrm{5}\:{digit}\:{number}. \\ $$$${what}'{s}\:{the}\:{probability}\:{that}\:{this}\:{number} \\ $$$${has}\:{exactly}\:\mathrm{3}\:{different}\:{digits}? \\ $$
Answered by mr W last updated on 03/Mar/23
from 10000 to 99999 there are totally 90000 5 digit numbers. now we want to find how many among them have exactly 3 digits. to select 3 digits from 0 to 9 there are C_3 ^(10) possibilities. say a, b, c are the digits we have selected. to form a 5 digit number using them we have following possibilities: 1) aaabc ⇒C_1 ^3 ×((5!)/(3!))=60 numbers 2) aabbc ⇒C_1 ^3 ×((5!)/(2!2!))=90 numbers in sum we have 150 numbers using a,b,c ⇒150×C_3 ^(10) =18000 numbers but among them some begin with the digit 0 like 0abab. to select a, b from 1 to 9 there are C_2 ^9 possibilities. 1) 0aaab ⇒2×((4!)/(3!))=8 numbers 2) 0aabb ⇒((4!)/(2!2!))=6 numbers 3) 000ab ⇒((4!)/(2!))=12 numbers 4) 00aab ⇒2×((4!)/(2!))=24 numbers in sum there are (8+6+12+24)×C_2 ^9 =1800 numbers which begin with 0. so we have 18000−1800=16200 numbers with exactly 3 digits. probability p=((16200)/(90000))=18% ✓
$${from}\:\mathrm{10000}\:{to}\:\mathrm{99999}\:{there}\:{are}\:{totally} \\ $$$$\mathrm{90000}\:\mathrm{5}\:{digit}\:{numbers}. \\ $$$${now}\:{we}\:{want}\:{to}\:{find}\:{how}\:{many}\:{among} \\ $$$${them}\:{have}\:{exactly}\:\mathrm{3}\:{digits}. \\ $$$${to}\:{select}\:\mathrm{3}\:{digits}\:{from}\:\mathrm{0}\:{to}\:\mathrm{9}\:{there}\:{are} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{10}} \:{possibilities}.\:{say}\:{a},\:{b},\:{c}\:{are}\:{the} \\ $$$${digits}\:{we}\:{have}\:{selected}.\:{to}\:{form}\:{a}\:\mathrm{5} \\ $$$${digit}\:{number}\:{using}\:{them}\:{we}\:{have} \\ $$$${following}\:{possibilities}: \\ $$$$\left.\mathrm{1}\right)\:{aaabc} \\ $$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{3}} ×\frac{\mathrm{5}!}{\mathrm{3}!}=\mathrm{60}\:{numbers} \\ $$$$\left.\mathrm{2}\right)\:{aabbc} \\ $$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{3}} ×\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{90}\:{numbers} \\ $$$${in}\:{sum}\:{we}\:{have}\:\mathrm{150}\:{numbers}\:{using} \\ $$$${a},{b},{c} \\ $$$$\Rightarrow\mathrm{150}×{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{18000}\:{numbers} \\ $$$${but}\:{among}\:{them}\:{some}\:{begin}\:{with}\:{the} \\ $$$${digit}\:\mathrm{0}\:{like}\:\mathrm{0}{abab}. \\ $$$${to}\:{select}\:{a},\:{b}\:{from}\:\mathrm{1}\:{to}\:\mathrm{9}\:{there}\:{are} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{9}} \:{possibilities}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{0}{aaab} \\ $$$$\Rightarrow\mathrm{2}×\frac{\mathrm{4}!}{\mathrm{3}!}=\mathrm{8}\:{numbers} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{0}{aabb} \\ $$$$\Rightarrow\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{6}\:{numbers} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{000}{ab} \\ $$$$\Rightarrow\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{12}\:{numbers} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{00}{aab} \\ $$$$\Rightarrow\mathrm{2}×\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{24}\:{numbers} \\ $$$${in}\:{sum}\:{there}\:{are}\: \\ $$$$\left(\mathrm{8}+\mathrm{6}+\mathrm{12}+\mathrm{24}\right)×{C}_{\mathrm{2}} ^{\mathrm{9}} =\mathrm{1800}\:{numbers} \\ $$$${which}\:{begin}\:{with}\:\mathrm{0}. \\ $$$${so}\:{we}\:{have}\:\mathrm{18000}−\mathrm{1800}=\mathrm{16200}\: \\ $$$${numbers}\:{with}\:{exactly}\:\mathrm{3}\:{digits}. \\ $$$${probability}\:{p}=\frac{\mathrm{16200}}{\mathrm{90000}}=\mathrm{18\%}\:\checkmark \\ $$
Commented by mr W last updated on 03/Mar/23
shorter way: 5 digit numbers with exact 3 different digits: (9/(10))×C_3 ^(10) ×3!× { (5),(3) :}} =(9/(10))×120×6×25=16200
$${shorter}\:{way}: \\ $$$$\mathrm{5}\:{digit}\:{numbers}\:{with}\:{exact}\:\mathrm{3}\:{different} \\ $$$${digits}: \\ $$$$\left.\frac{\mathrm{9}}{\mathrm{10}}×{C}_{\mathrm{3}} ^{\mathrm{10}} ×\mathrm{3}!×\begin{cases}{\mathrm{5}}\\{\mathrm{3}}\end{cases}\right\} \\ $$$$=\frac{\mathrm{9}}{\mathrm{10}}×\mathrm{120}×\mathrm{6}×\mathrm{25}=\mathrm{16200} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *