(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
Answer:
11,15,19,23,27,31,35,39,43,47,51,....
Answer:
sin I = 
Step-by-step explanation:
sin I = 
= 
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Step-by-step explanation:
I will solve your system by substitution.
(You can also solve this system by elimination.)
4x+y=17;4x−3y=−3
Step: Solve4x+y=17for y:
4x+y+−4x=17+−4x(Add -4x to both sides)
y=−4x+17
Step: Substitute−4x+17foryin4x−3y=−3:
4x−3y=−3
4x−3(−4x+17)=−3
16x−51=−3(Simplify both sides of the equation)
16x−51+51=−3+51(Add 51 to both sides)
16x=48
16/x = 48/16
(Divide both sides by 16)
x=3
Step: Substitute 3 for x in y=−4x+17:
y=−4x+17
y=(−4)(3)+17
y=5(Simplify both sides of the equation)
Answer:
x=3 and y=5
