Clear way please How many ways are there to pick a man and a woman who are not h
ID: 3852800 • Letter: C
Question
Clear way please How many ways are there to pick a man and a woman who are not husband and wife from a group of n married couples? How many nonempty words can be formed from three As and five Bs? (not all letters must be used, any sequence of letters counts as a word) How many ternary (0, 1, 2) sequences of length 10 are there without any two consecutive digits being the same? (Ternary means using digits 0, 1, 2. Similarly, binary would mean just digits 0, 1. No consecutive the same means 11, 22,00 are the forbidden substrings.) How many different outcomes are possible when a pair of dices, one red and one white are rolled two consecutive times? Consider that one roll consists of outcome that is formed by the PAIR of what is on white and red and count the number of outcomes if A) it is possible to distinguish which roll was first and which was second. B) it is not possible to distinguish first and second roll. (Consider that first and second roll are distinguishable as well as the case where they are not. Note that one roll is. which is red and white dices is equivalent to a roll of a dice with 36 sides.) How many integral solutions of x_1 + x_2 + x_3 + x_4 = 30 satisfy x_1 greaterthanorequalto 2, x_2 greaterthanorequalto 0, x_3 greaterthanorequalto - 5, and x_4 greaterthanorequalto 8? (Use substitution to get greaterthanorequalto for all variables.)Explanation / Answer
1.) there are N men and N woman
Ways to choose a man =N
Ways to choose a women=N-1, therefore answer
[1Cn][1C(n-1)] = n!/{1![n-1]!} * {[n-1]!/1![n-2]!}
Answer=n[n-1]
2.)One item can be selected in 2 ways.
two items can be selected in 3 ways.
Three item can be selected in 4 ways.
Five item can be selected in 4 ways.
Six item can be selected in 3 ways.
seven items can be selected in 2 ways.
eight items can be selected in 1 way.
Total: 2×(2+3+4)+4+1=232×(2+3+4)+4+1=23 ways
3.) there are 10 spaces which can be filled with 0,1 and 2.
Now the first space can be filled with 3 ways and as the digits can't be consecutive, therefore next space will filled with 2 ways and also the last 8 spaces can be 2 ways.
So the answer will be 3*2^9= 1536.
4.)when both dice are rolled first time total outcome=36
Similarly for second time outcome =36
Therefore total outcome for two consecutive time=>36*36=1296
A) When roll is distinguishable with 36 sided dice then in first and second roll we get 36*36 i.e.6^4.
B) When indistinguishable, by analogy with two identical dice we can draw a table 6x6 with possible outcome.There are 6 doubles, and the remaining (6^2 - 6) outcomes come in pairs: (2,3) is indistinguishable from (3,2), we can half (6^2 -6), and add 6 in doubles.
Similarly with 36-sided dice, we can get n + (1/2)(n^2 - n) = (1/2)(n^2 + n)
i.e. 36+(1/2)(36^2-36)outcomes, when n = 36 which is 18*37=666.
5.)Let y1 = x1 2,
y2 = x2,
y3 = x3 + 5,
y4 = x4 8.
Now the integral solution of x1 + x2 + x3 + x4 = 30 where x1 2, x2 0, x3 5 and x4 8 corresponds to an integral solution of y1 + y2 + y3 + y4 = 25 such that y1,... ,y4 0.
From a result in class,
|{(y1,y2,y3,y4) : y1,... ,y4 Z+ and y1 + · · · + y4 = 25}| =
(25 + 4 1)C(4 1) =28C3 =3276
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