(a) how many positive integers are there between 1000 and 4999, inclusive?

(b) how many positive integers between 1000 and 4999, inclusive:

1. have no repeated digit?

2. have at least one repeated digit?

3. have at most two repeated digits?

note that by `one repeated digit' we mean that there is a digit that appears

at least twice (eg, 1123 has one repeated digit). similarly, by `two repeated digits'

we mean a digit that appears at least three times (eg, 1112).

a) 4000

b1) 480

b2) 3,520

b3) 1,434

Step-by-step explanation:

a)

4999 - 1000 +1 = 4000

b)

We must find numbers of the form “abcd” where “a” is 1,2,3 or 4 and all the digits are different to each other.

The digits “bcd” must be chosen from the set {0,5,6,7,8,9} with no repetition. So by the fundamental rule of counting, we have

4*6*5*4 = 480 numbers between 1000 and 4999 with no repeated digits.

2.

We can see that this set is the complement of the previous one, so there are 4,000 - 480 = 3,520

3.

Let's compute the amount of number with exactly 2 repeated digits and add it to 480 (amount with no repeated digits)

Starting with 1

we have numbers of the form 11xy, 1x1y and 1xy1 where xy are permutations of the set {0,2,3,4,5,6,7,8,9} taken 2 at a time, so we have 72+72+72 = 216.

The same procedure is repeated for 2,3 and 4 and we have  

216+216+216 +216 = 864 numbers starting with 1,2,3 or 4 and exactly two repeated digits.

Now we found the triplets xxy, xyx, yxx with exactly 2 repeated digits none of them 1,2,3 or 4 and we have  

15+15+15+15+15+15 = 90

And finally we have

864+90+480 = 1,434 digits with at most 2 repeated digits.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.