3 ones = C(5, 3)

Home Work 7 (Discrete)
1- How many binary strings of length 5 contain at least 3 ones?
-
3 ones = C(5, 3)
4 ones = C(5, 4)
5 ones = C(5, 5)
5
5
5
5
5
βˆ‘ ( ) = ( ) + ( ) + ( ) = 10 + 5 + 1 = 16
𝑛
3
4
5
𝑛=3
2- How many integer numbers between 100 and 1000 have digits in decreasing order?
All integer numbers from 100 to 1000 consist of three digits, so we are looking for
three digit numbers have their digits in decreasing order such as 321.
(10
) = 120
3
10βˆ—9βˆ—8
OR
=
3!
720
6
= 120
3- Find the number of integer solutions of equation x1 + x2 + x3 = 12 where
π‘₯1 β‰₯ 2, π‘₯2 β‰₯ 3, π‘₯3 β‰₯ 4
Replace variables of the equation:
x1 βˆ’ 2 = π‘₯́ 1
x2 βˆ’ 3 = π‘₯́ 2
x3 βˆ’ 4 = π‘₯́ 3
Then find the number of nonnegative solutions of equation
π‘₯́ 1 + π‘₯́ 2 + π‘₯́ 3 = 3
(
π‘€β„Žπ‘’π‘Ÿπ‘’ π‘₯́ 1 β‰₯ 0 π‘₯́ 2 β‰₯ 0 π‘₯́ 3 β‰₯ 0
3+3βˆ’1
5
) = ( ) = 10 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›π‘ 
3βˆ’1
2
4- For given expression (π‘₯ + 𝑦)13
a. Find the number of elements of expansion
𝑛+π‘šβˆ’1
(
) π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑛 = 13 (π‘‘β„Žπ‘’ π‘π‘œπ‘€π‘’π‘Ÿ) π‘Žπ‘›π‘‘ π‘š = 2 (π‘₯&𝑦)
π‘šβˆ’1
(
13 + 2 βˆ’ 1
14
) = ( ) = 14
2βˆ’1
1
b. Find the coefficient of element π‘₯ 5 𝑦 8
(
13
) = 1287
8
5- How many words with 8 letters can be created from letters F,G,H,I,J
a) That contains substring FG, GH or both?
|𝐴 βˆͺ 𝐡| = |𝐴| + |𝐡| βˆ’ |𝐴 ∩ 𝐡|
7
7
7
( ) 56 + ( ) 56 βˆ’ (( ) βˆ— 54 ) =
1
1
2
7 βˆ— 15625 + 7 βˆ— 15625 βˆ’ 21 βˆ— 625 = 109375 + 109375 βˆ’ 13125 = 205625
b) Where H appears before I?
(82)
2
βˆ— 56 =
28
βˆ— 15625 = 218750
2
6- How many strings of 8 ternary digits (0, 1, or 2) is there that contain exactly two 0s,
three 1s, and three 2s?
C(8,2) * C(6,3) * C(3,3) = 560
7- Find the number of binary strings of length 5 that contain substring β€˜011’ or β€˜110’.
|𝐴 βˆͺ 𝐡| = |𝐴| + |𝐡| βˆ’ |𝐴 ∩ 𝐡|
(31)22 + (31)22 βˆ’ 0 = 12 + 12 – 0 = 24
8- Find the number of binary strings of the length 8 that contain 5 zeroes, 3 ones, and do
not contain substring "11". Give the reason of your solution
Result = (ALL binary strings of the length 8 that contain 5 zeroes 3 ones) – (Binary
strings of the length 8 that contain 5 zeroes, 3 ones, and contain substring "11")
C(8,5) * C(3,3) – C(7,1) * C(6,1) * 1 =
56 * 1 – 7 * 6 * 1 = 56 – 42 = 14