Probability (1 Viewer)

[JustRandomThings]

Hey all thanks in advance

Consider the digits 9,8,7,6,5,4,3,2,1,0. Find how many five-digit numbers are possible if the digits are to be in:

a) descending order

b) ascending order

Combination doesn't seem to work here... either that or I'm thinking wrong ;p

 

[pikachu975]

a) 6x5x4x3x2 = 720
b) 6x5x5x3x2 = 720

lol probably completely wrong not that good at perms and combs.

 

[JustRandomThings]

a) 6x5x4x3x2 = 720
b) 6x5x5x3x2 = 720

lol probably completely wrong not that good at perms and combs.

Lol the answer is 252 and 126 ^^

 

[InteGrand]

For descending order:

Case 1. Our quintuple of digits contains a 0.

The 0 has to go at the end, and we need to pick four more numbers from nine available and place them in descending order. This is equal to the no. of ways to just select four numbers, since each selection has precisely one way of ordering them in descending order. So there are 9C4 ways for this case.

Case 2. No 0 picked.

By similar reasoning, 9C5 ways here.

So the answer is 9C4 + 9C5 = 252.

 

[InteGrand]

For ascending order, there is a one-to-one correspondence with the descending order permutations (by flipping the digits around), except that the ones in the descending order with 0 at the end (Case 1 above) won't actually count since that'd give us a permutation with 0 at the start, which we don't want when making a five digit number. So for this one, the answer is just the ways from Case 2 above, which is half the previous answer, i.e. 126.

 

[pikachu975]

Lol the answer is 252 and 126 ^^

Rip perms and combs lol