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=The Zeta Monogram= | =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 + ... | ||
Revision as of 20:47, 16 November 2019
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?
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 + ...
Using the formula n(n+1)/2 we can see that 24 + 23 + ... = 24*25/2 = 300