Discrete Maths (Degree level)

Hi
I have been set some more tricky questions, please can anyone help?

1. (i) How many three-digit integers <= 1000 have their digits increasing from left to right (like 289)
How many have their digits decreasing?
(ii) How many three-digit integers <= 1000000 have their digits increasing from left to right

2.
(i) In how many different ways can we choose a 'hand' of 2 cards from a pack of 52 cards
(ii) in how many different ways can we choose 2 cards of different suits?
(iii) in how many different ways can we choose 2 cards with different 'face values' (such as a 2 and a 7 or a 9 and a 3)

3. 20 children have to line up for a class photo. Given that Joe refuses to stand next to Lucy, how many possible orders are there?

4. How many integers between 1 and 1000 have exactly one 6 in them?

Thanks very much

Reply 1

Jonny W

(n choose m) means n!/[m!(n - m)!].

(1)
(i) None of the digits can be 0. To get one of the required integers we choose 3 digits from {1, ..., 9} and arrange them in ascending order. Number of ways of choosing = (9 choose 3) = 84.

Turning to the second part, to get one of the required integers we choose 3 digits from {0, ..., 9} and arrange them in descending order. Number of ways of choosing = (10 choose 3) = 120.

(ii) [I guess the question should say "six-digit" rather than "three-digit".] Answer = (9 choose 6) = 84.

(2)
(i)
(52 choose 2) = 1326

(ii)
Number of ways of choosing one club card and one spade card
= 13^2
= 169

Answer
= Number of ways of choosing two suits * 169
= (4 choose 2) * 169
= 1014

(iii)
Answer
= (52 choose 2) - Number of ways of choosing 2 cards with the same face value
= (52 choose 2) - 13*(4 choose 2)
= 1248

(3)
There are 20! ways of arranging the children.

Of these, there are 19! orders in which Joe stands immediately to the left of Lucy. (Imagine tying Joe and Lucy together.) There are also 19! orders in which Joe stands immediately to the right of Lucy.

Hence there are 2*19! orders in which Joe stands next to Lucy.

So there 20! - 2*19! = 18*19! orders in which Joe does not stand next to Lucy.

(4)
We start by counting the three-digit numbers with exactly one 6. If the 6 is the first digit then there are 9^2 ways of choosing the other digits. If the 6 is the second or third digit then there are 8*9 ways of choosing the other digits (because the first digit can't be 0). So there are 9^2 + 2*8*9 = 225 three-digit numbers with exactly one 6.

Similarly there 9 + 8 = 17 two-digit numbers with exactly one 6.

There is 1 one-digit number with exactly one 6.

Answer = 225 + 17 + 1 = 243

[Alternative]