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How many different positive three-digit integers can be formed using only the digits in the set \(\{2, 3, 5, 5, 5, 6, 6\}\)if no digit may be used more times than it appears in the given set of available digits?

 Jan 18, 2021
 #1
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+1

2356    choose 3      12 ways

   now suppose there are two 5's    there are 3 ways to do this...the remaining spot can be 23 or 6    3 * 3

       suppose there are three 5's      one way

   now suppose there are two 6's    there are three wys to place them...remaining spot can be 2 3 or 5    3 * 3

 

12 + 3*3  + 1 + 3* 3 =   31 numbers    ???      Just my guess....  Maybe MMM will do better/correctly !  

 Jan 18, 2021
 #4
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+1

*** Updated ***

 

2356       4 x 3 x 2  = 24 numbers     (this is 4 P 3 )

  now suppose there are two 5's    there are 3 ways to do this...the remaining spot can be 23 or 6    3 * 3

       suppose there are three 5's      one way

   now suppose there are two 6's    there are three wys to place them...remaining spot can be 2 3 or 5    3 * 3

 

24 + 3*3  + 1 + 3* 3 =  43 numbers    ???         *** edited ***

Guest Jan 19, 2021
 #2
avatar+536 
+1

235

236

253

263

255

266

265

256

Since there is also only 1 3 there are also 8 numbers that start at 3.

So we have 16 so far

For 6 we have

623

632

625

652

635

653

655

626

662

636

663

665

656

Then for 5 we have

523 

532

525

552

535

553

565

556

566

526

536

563

562

555

So I think that's it, 16+ the 6 column+ the 5 column

 Jan 19, 2021
 #3
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+1

[(2, 3, 5), (2, 3, 6), (2, 5, 5), (2, 5, 6), (2, 6, 6), (3, 5, 5), (3, 5, 6), (3, 6, 6), (5, 5, 5), (5, 5, 6), (5, 6, 6)],  >Total distinct combinations = 11

 

Convert the above combinations into permutations as follows:


[6 + 6 + 3 + 6 + 3 + 3 + 6 + 3 + 1 + 3 + 3] = 43 permutations. They can be listed as follows:

 

[(2, 3, 5), (2, 3, 6), (2, 5, 3), (2, 5, 5), (2, 5, 6), (2, 6, 3), (2, 6, 5), (2, 6, 6), (3, 2, 5), (3, 2, 6), (3, 5, 2), (3, 5, 5), (3, 5, 6), (3, 6, 2), (3, 6, 5), (3, 6, 6), (5, 2, 3), (5, 2, 5), (5, 2, 6), (5, 3, 2), (5, 3, 5), (5, 3, 6), (5, 5, 2), (5, 5, 3), (5, 5, 5), (5, 5, 6), (5, 6, 2), (5, 6, 3), (5, 6, 5), (5, 6, 6), (6, 2, 3), (6, 2, 5), (6, 2, 6), (6, 3, 2), (6, 3, 5), (6, 3, 6), (6, 5, 2), (6, 5, 3), (6, 5, 5), (6, 5, 6), (6, 6, 2), (6, 6, 3), (6, 6, 5)]   >Total distinct permutations = 43

 Jan 19, 2021
 #5
avatar+112523 
+1

 

How many different positive three-digit integers can be formed using only the digits in the set {2,3,5,5,5,6,6} if no digit may be used more times than it appears in the given set of available digits?

 

555                                1

55 and one other    3*3=9

66  and one other    3*3=9

all different              4*3*2=24

 

24+9+9+1 = 43

 Jan 24, 2021

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