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Semenov [28]
2 years ago
11

A small combination lock on a suitcase has 4 ​wheels, each labeled with the 10 digits 0 to 9. how many 4 digit combinations are

possible if no digit is​ repeated? if digits can be​ repeated? if successive digits must be​ different?
Mathematics
1 answer:
tamaranim1 [39]2 years ago
8 0
This is the easiest way to solve this problem:

Imagine this represents how many combinations you can have for each of the 4 wheels (each blank spot for one wheel): __ __ __ __

For the first situation it says how many combos can we make if no digits are repeated.
We have 10 digits to use for the first wheel so put a 10 in the first slot 
10 __ __ __
Since no digit can be repeated we only have 9 options for the second slot
10 9_ __ __
Same for the third slot, so only 8 options
<u>10</u> <u> 9 </u> <u> 8 </u> __
4th can't be repeated so only 7 options left
<u>10</u> <u> 9 </u> <u> 8 </u> <u> 7 
</u><u>
</u>Multiply the four numbers together: 10*9*8*7 = 5040 combinations


For the next two do the same process as the one above.

If digits can be repeated? You have ten options for every wheel so it would look like this: <u>10</u> <u>10</u> <u>10</u> <u>10
</u>
10*10*10*10 = 10,000 combinations

If successive digits bust be different?
We have 10 for the first wheel, but second wheel only has 9 options because 2nd number can't be same as first. The third and fourth wheels also has 9 options for the same reason.

<u>10</u> <u> 9</u><u> </u> <u> 9 </u> <u> 9 </u>

10*9*9*9 = 7290 combinations






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Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15
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\huge \boxed{\mathfrak{Question} \downarrow}

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\left. \begin{array}  { l  }  { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.

  • To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

2x-5y=-13, \: 3x+4y=15

  • Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

2x-5y=-13

  • Add 5y to both sides of the equation.

2x=5y-13

  • Divide both sides by 2.

x=\frac{1}{2}\left(5y-13\right)  \\

  • Multiply \frac{1}{2}\\ times 5y - 13.

x=\frac{5}{2}y-\frac{13}{2}  \\

  • Substitute \frac{5y-13}{2}\\ for x in the other equation, 3x + 4y = 15.

3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15  \\

  • Multiply 3 times \frac{5y-13}{2}\\.

\frac{15}{2}y-\frac{39}{2}+4y=15  \\

  • Add \frac{15y}{2} \\ to 4y.

\frac{23}{2}y-\frac{39}{2}=15  \\

  • Add \frac{39}{2}\\ to both sides of the equation.

\frac{23}{2}y=\frac{69}{2}  \\

  • Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

\large \underline{ \underline{ \sf \: y=3 }}

  • Substitute 3 for y in x=\frac{5}{2}y-\frac{13}{2}\\. Because the resulting equation contains only one variable, you can solve for x directly.

x=\frac{5}{2}\times 3-\frac{13}{2}  \\

  • Multiply 5/2 times 3.

x=\frac{15-13}{2}  \\

  • Add -\frac{13}{2}\\ to \frac{15}{2}\\ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

\large\underline{ \underline{ \sf \: x=1 }}

  • The system is now solved. The value of x & y will be 1 & 3 respectively.

\huge\boxed{  \boxed{\bf \: x=1, \: y=3 }}

8 0
2 years ago
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