# Brooklyn Tech License Plates

by Andrew
(New York)

The Brooklyn Tech Alumni Foundation is producing a set of commemorative license plates. Each plate contains a sequence of 5 characters. Each character is chosen from the four letters in TECH and the four digits in 2012. No character may appear more times on the license plate than it appears among the four letters in TECH or the four digits in 2012 (i.e. a 2 may appear, at most, twice, but all other characters can only appear, at most, once). How many different license plates are possible?

### Comments for Brooklyn Tech License Plates

 May 16, 2012 Maximum Number of License Plates by: Staff Part I Question: by Andrew (New York) The Brooklyn Tech Alumni Foundation is producing a set of commemorative license plates. Each plate contains a sequence of 5 characters. Each character is chosen from the four letters in TECH and the four digits in 2012. No character may appear more times on the license plate than it appears among the four letters in TECH or the four digits in 2012 (i.e. a 2 may appear, at most, twice, but all other characters can only appear, at most, once). How many different license plates are possible? Answer: Only 7 unique alpha/number characters that can be stamped on the license plates. 2 is the only character that can repeated twice: TECH 201 Two calculations are necessary to answer this question: 1) How many different alpha/number sequences are possible if the number “2” is used no more than once (either 0 or 1 time) on each license plate? The order of the alpha/number sequence for each license plate must be different. However, the characters are always selected from the same 7 possibilities. Because the order of the alpha/number sequences is important, we will calculate the number of “permutations” possible using the seven possibilities (TECH 201), selected 5 at a time. The formula for this calculation is: Order is important No repetition of the same combination Permutations possible = n_P_r = n!/[(n-r)!] n (number of items to choose from) = 7 r (number of choices = 5 Permutations possible = 7!/[(7-5)!] = 7!/[(7-5)!] = (7*6*5*4*3*2*1)/(2*1) = 2520 possible alpha/number sequences ----------------------------------------------------------------

 May 16, 2012 Maximum Number of License Plates by: Staff ---------------------------------------------------------------- Part II 2) How many additional alpha/number sequences are possible if the number “2” is used twice on some of the license plates? If the “2” takes up two out of five places in the 5 character sequence, the number of permutations for the remaining three spaces is: Order is important No repetition of the same character/number n (number of items to choose from) = 6 r (number of choices) = 3 Permutations possible = n_P_r = n!/[(n-r)!] = 6!/[(6-3)!] = 6!/[(3)!] = (6*5*4*3*2*1)/(3*2*1) = 120 possible alpha/number sequences for 3 characters (which are not “2”) Every one of these 120 possible 3 character sequences leaves two places for two 2’s Now we are going to calculate how many different ways two 2’s can be inserted in the 120 three character sequences. But first we are going to calculate how many ways two 2’s can be inserted in just one of the three character sequences (which are actually 5 character sequences with two blanks for the 2’s) The formula for this calculation is NOT the same as we used in the first calculation: In this instance, we care going to calculate the number of combinations, rather than the number of permutations.. ORDER is NOT important REPETITION NOT ALLOWED. Combinations possible = n_C_r = n!/[r!(n-r)!] n = 5 r = 2 Combinations possible = n_C_r = 5!/[2!(5-2)!] = 5!/[2!(5-2)!] = (5*4*3*2*1)/[2*1*(3*2*1)] = (5*4)/(2*1) = (20)/(2) = 10 The “2’s” can be added to every one of the 3 character sequences in 10 different ways. These are: 2,2,x,x,x; 2,x,2,x,x; 2,x,x,2,x; 2,x,x,x,2; x,2,x,x,2; x,x,2,x,2; x,x,x,2,2; x,x,2,2,x; x,2,2,x,x; x,2,x,2,x Since there are 120 possible 3 character sequences, the total number of 5 character sequences with two 2’s is: = 10 * 120 = 1200 The sum of all the possibilities = 1) + 2) N = 2520 + 1200 = 3720 >>> the final answer to the question is: 3720 different license plates are possible Thanks for writing. Staff www.solving-math-problems.com

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