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Amrabdelnaby
- Master | Next Rank: 500 Posts
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Dear Pros,
Pls tell which part of my calculation is wrong in answering the following question:
Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?
90
82
80
45
36
Here is how i approached the problem:
we will have three cases
s s d
s d s
d s s where s represents same and d represents different
for the first case (s s d) we start with the most restrictive digit, the first one. we will have 3 options for this digit, either 7, 8, or 9
as for the third digit, which is restrictive also and must be different from the first, we can't use 0 as the number must be greater than 700 and we can't use the number in the first digit, so we have 8 options. and for the middle digit we have only 1 option for it to be similar to the last digit. so for the first case we have 3x1x8 = 24 ways
for the second case (s d s) first digit will have 3 options, last has 1 option and the middle has 9 options, so we have 3x9x1 = 27 ways
for the last case (d s s ) first digit will have 3 options, last will have 8 as it must not include a zero nor the number used in the first digit, and the middle digit will have one option giving us 3x1x8 = 24
adding the three cases we get 24+27+24 = 75 different ways!
this answer is not even in the answer choices so I must have done something wrong.
pls enlighten me
Pls tell which part of my calculation is wrong in answering the following question:
Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?
90
82
80
45
36
Here is how i approached the problem:
we will have three cases
s s d
s d s
d s s where s represents same and d represents different
for the first case (s s d) we start with the most restrictive digit, the first one. we will have 3 options for this digit, either 7, 8, or 9
as for the third digit, which is restrictive also and must be different from the first, we can't use 0 as the number must be greater than 700 and we can't use the number in the first digit, so we have 8 options. and for the middle digit we have only 1 option for it to be similar to the last digit. so for the first case we have 3x1x8 = 24 ways
for the second case (s d s) first digit will have 3 options, last has 1 option and the middle has 9 options, so we have 3x9x1 = 27 ways
for the last case (d s s ) first digit will have 3 options, last will have 8 as it must not include a zero nor the number used in the first digit, and the middle digit will have one option giving us 3x1x8 = 24
adding the three cases we get 24+27+24 = 75 different ways!
this answer is not even in the answer choices so I must have done something wrong.
pls enlighten me
