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# helllllppppp

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Find the number of positive integers that satisfy both the following conditions: Each digit is a 1 or a 2 The sum of the digits is 8

Mar 31, 2023

#1
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We can solve this problem by using casework and counting the number of possibilities for each case.

Case 1: The number has one digit that is equal to 1 and the other digits are equal to 2. There are 8 ways to choose the position of the digit that is equal to 1 (it can be any of the 8 digits). Once we have chosen the position of the digit that is equal to 1, there are 2 possibilities for each of the other digits, for a total of \$2^7\$ possibilities.

Case 2: The number has two digits that are equal to 1 and the other digits are equal to 2. There are \$\binom{8}{2} = 28\$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of \$2^6\$ possibilities.

Case 3: The number has three digits that are equal to 1 and the other digits are equal to 2. There are \$\binom{8}{3} = 56\$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of \$2^5\$ possibilities.

Case 4: The number has four digits that are equal to 1 and the other digits are equal to 2. There are \$\binom{8}{4} = 70\$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of \$2^4\$ possibilities.

Case 5: The number has five digits that are equal to 1 and the other digits are equal to 2. There are \$\binom{8}{5} = 56\$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of \$2^3\$ possibilities.

Case 6: The number has six digits that are equal to 1 and the other digits are equal to 2. There are \$\binom{8}{6} = 28\$ ways to choose the positions of the digits that are equal to 1. Once we have chosen the positions of the digits that are equal to 1, there are 2 possibilities for each of the other digits, for a total of \$2^2\$ possibilities.

Case 7: The number has seven digits that are equal to 1 and the other digit is equal to 2. There are 8 ways to choose the position of the digit that is equal to 2. Once we have chosen the position of the digit that is equal to 2, there are 2 possibilities for each of the other digits, for a total of \$2^7\$ possibilities.

Case 8: The number has eight digits that are equal to 2. There is only one possibility in this case.

Thus, the total number of positive integers that satisfy both conditions is:

2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^7 + 1 = 381.

Mar 31, 2023
#4
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Thanks!

Guest Apr 4, 2023
#2
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2222 , 11222 , 12122 , 12212 , 12221 , 21122 , 21212 , 21221 , 22112 , 22121 , 22211 , 111122 , 111212 , 111221 , 112112 , 112121 , 112211 , 121112 , 121121 , 121211 , 122111 , 211112 , 211121 , 211211 , 212111 , 221111 , 1111112 , 1111121 , 1111211 , 1112111 , 1121111 , 1211111 , 2111111 , 11111111 , Total =  34 such integers.

Mar 31, 2023
#5
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The correct answer is 381, but thanks for trying!

Guest Apr 4, 2023
#3
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(2222)==4! /4! ==1           permutations
(11222)==5! / 3!2!==10    p
(111122)==6! / 4!2!==15  p
(1111112)==7! / 6!==7      p
(11111111)==8! / 8!==1     p

Total ==1 + 10 + 15 + 7 + 1 ==34 number of positive integers.

Apr 1, 2023