Difference between revisions of "2020 AMC 8 Problems/Problem 7"
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− | ==Problem | + | ==Problem== |
− | How many integers between <math>2020</math> and <math>2400</math> have four distinct digits arranged in increasing order? (For example, <math>2347</math> is one integer.) | + | How many integers between <math>2020</math> and <math>2400</math> have four distinct digits arranged in increasing order? (For example, <math>2347</math> is one integer.). |
<math>\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}</math> | <math>\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}</math> | ||
==Solution 1== | ==Solution 1== | ||
− | + | Firstly, observe that the second digit of such a number cannot be <math>1</math> or <math>2</math>, because the digits must be distinct and increasing. The second digit also cannot be <math>4</math> as the number must be less than <math>2400</math>, so it must be <math>3</math>. It remains to choose the latter two digits, which must be <math>2</math> distinct digits from <math>\left\{4,5,6,7,8,9\right\}</math>. That can be done in <math>\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15</math> ways; there is then only <math>1</math> way to order the digits, namely in increasing order. This means the answer is <math>\boxed{\textbf{(C) }15}</math>. | |
− | |||
− | ==Solution 2= | + | ==Solution 2 (without using the "choose" function)== |
+ | As in Solution 1, we find that the first two digits must be <math>23</math>, and the third digit must be at least <math>4</math>. If it is <math>4</math>, then there are <math>5</math> choices for the last digit, namely <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math>. Similarly, if the third digit is <math>5</math>, there are <math>4</math> choices for the last digit, namely <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math>; if <math>6</math>, there are <math>3</math> choices; if <math>7</math>, there are <math>2</math> choices; and if <math>8</math>, there is <math>1</math> choice. It follows that the total number of such integers is <math>5+4+3+2+1=\boxed{\textbf{(C) }15}</math>. | ||
− | + | ==Video Solution by NiuniuMaths (Easy to understand!)== | |
+ | https://www.youtube.com/watch?v=8hgK6rESdek&t=9s | ||
− | + | ~NiuniuMaths | |
− | == | + | ==Video Solution by Math-X (First understand the problem!!!)== |
− | + | https://youtu.be/UnVo6jZ3Wnk?si=xBEcgPkL367f3Zp8&t=713 | |
− | + | ||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (🚀Fast🚀)== | ||
+ | https://youtu.be/QqBpLTQojHg | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/FjmBtgrGfCs | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/61c1MR9tne8 ~ The Learning Royal | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=251 | ||
− | + | ~Interstigation | |
− | |||
− | + | ==Video Solution by STEMbreezy== | |
+ | https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85 | ||
+ | ~STEMbreezy | ||
− | + | ==See also== | |
+ | {{AMC8 box|year=2020|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:02, 12 June 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (without using the "choose" function)
- 4 Video Solution by NiuniuMaths (Easy to understand!)
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (🚀Fast🚀)
- 7 Video Solution by WhyMath
- 8 Video Solution
- 9 Video Solution by Interstigation
- 10 Video Solution by STEMbreezy
- 11 See also
Problem
How many integers between and
have four distinct digits arranged in increasing order? (For example,
is one integer.).
Solution 1
Firstly, observe that the second digit of such a number cannot be or
, because the digits must be distinct and increasing. The second digit also cannot be
as the number must be less than
, so it must be
. It remains to choose the latter two digits, which must be
distinct digits from
. That can be done in
ways; there is then only
way to order the digits, namely in increasing order. This means the answer is
.
Solution 2 (without using the "choose" function)
As in Solution 1, we find that the first two digits must be , and the third digit must be at least
. If it is
, then there are
choices for the last digit, namely
,
,
,
, or
. Similarly, if the third digit is
, there are
choices for the last digit, namely
,
,
, and
; if
, there are
choices; if
, there are
choices; and if
, there is
choice. It follows that the total number of such integers is
.
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=xBEcgPkL367f3Zp8&t=713
~Math-X
Video Solution (🚀Fast🚀)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution
https://youtu.be/61c1MR9tne8 ~ The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=251
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85
~STEMbreezy
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.