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=The Zeta Monogram= | {{FMM | ||
|title=The Zeta Monogram | |||
|problem= | |||
Mrs. and Mrs. Zeta are expecting a new (gender-neutral) baby. They wish to name their baby such that the baby's monogram (first initial, middle initial, last initial) is in alphabetical order with no letter repeated. (For example, ABZ is a valid monogram, but BBZ and QBZ are not.) How many possible monograms are there? | Mrs. and Mrs. Zeta are expecting a new (gender-neutral) baby. They wish to name their baby such that the baby's monogram (first initial, middle initial, last initial) is in alphabetical order with no letter repeated. (For example, ABZ is a valid monogram, but BBZ and QBZ are not.) How many possible monograms are there? | ||
|solution= | |||
The Zeta family wants little baby Zeta to have a monogram (first initial, middle initial, last initial) that is alphabetical with no letters repeated. The last letter of the monogram must be Z. If the first letter is A, that leaves 26 - 2 = 24 possible letters, ABZ, ACZ, etc. If the first letter is B, that leaves 23 possible letters, BCZ, BDZ, etc. If we repeat this for all possibilities we will get 24 + 23 + 22 + ... | The Zeta family wants little baby Zeta to have a monogram (first initial, middle initial, last initial) that is alphabetical with no letters repeated. The last letter of the monogram must be Z. If the first letter is A, that leaves 26 - 2 = 24 possible letters, ABZ, ACZ, etc. If the first letter is B, that leaves 23 possible letters, BCZ, BDZ, etc. If we repeat this for all possibilities we will get 24 + 23 + 22 + ... | ||
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Using the formula n(n+1)/2 we can see that 24 + 23 + ... = 24*25/2 = 300 | Using the formula n(n+1)/2 we can see that 24 + 23 + ... = 24*25/2 = 300 | ||
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Latest revision as of 22:09, 16 November 2019
Friday Morning Math Problem
The Zeta Monogram
Mrs. and Mrs. Zeta are expecting a new (gender-neutral) baby. They wish to name their baby such that the baby's monogram (first initial, middle initial, last initial) is in alphabetical order with no letter repeated. (For example, ABZ is a valid monogram, but BBZ and QBZ are not.) How many possible monograms are there?
| Solution |
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The Zeta family wants little baby Zeta to have a monogram (first initial, middle initial, last initial) that is alphabetical with no letters repeated. The last letter of the monogram must be Z. If the first letter is A, that leaves 26 - 2 = 24 possible letters, ABZ, ACZ, etc. If the first letter is B, that leaves 23 possible letters, BCZ, BDZ, etc. If we repeat this for all possibilities we will get 24 + 23 + 22 + ...
Using the formula n(n+1)/2 we can see that 24 + 23 + ... = 24*25/2 = 300
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Friday Morning Math Problems weekly math problems
Spider Socks and Shoes FMM1 Sums of Powers of 2 FMM2 Fifty Coins FMM3 The Zeta Monogram FMM4 The Cthulhus Monogram FMM4B Multiplication Logic FMM5 The Termite and the Cube FMM6 Sharing Dump Trucks FMM7 The Flippant Juror FMM8 Bus Routes FMM9 A Robust Bus System FMM9B Square-Free Sequence FMM10 Inferring Rule from Sequence FMM11 Checkerboard Color Schemes FMM12 One-Handed Chords FMM13 First Ace FMM14 Which Color Cab FMM15 Petersburg Paradox Revisited FMM16 A Binomial Challenge FMM17 A Radical Sum FMM18 Memorable Phone Numbers FMM19 Arrange in Order FMM20 A Pair of Dice Games: FMM21
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