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?

Guest Jan 18, 2021

#1**+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 !

Guest Jan 18, 2021

#4**+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**+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

MooMooooMooM Jan 19, 2021

#3**+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**

Guest Jan 19, 2021

#5**+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**

Melody Jan 24, 2021