Answer to Question #187958 in Discrete Mathematics for Angelie Suarez

Question

Solve the following. (mod4)

1. How many 3-digit number can be formed from digits 1 – 5 if:

a. If repetition is allowed?

b. If repetition is not allowed?

2. How many strings of length 4 can be formed using the letters

COMPUTER if repetitions are not allowed?

3. Out of 17 regions in the country, ten will be chosen to be included

in a survey. How many ways of selecting 10 out of 17 regions?

4. Expand the (2𝑥 − 𝑦)3 using binomial coefficient.

5. Find for the coefficient of the following:

a. x4y3 after the expansion of (x – 2y )7

b. x6y6 after the expansion of (2x + y )12

6. Find the 9th term of the expression (2x + y )12

Answer 1

1(a) A three digit number is to be formed from given 5 digits 1,2,3,4,5.

Now, there are 5 ways to fill ones place.

Since, repetition is allowed , so tens place can be filled by all 5 digits.

and tens place can be filled in 5 ways.

Similarly, to fill hundreds place, we have also 5 digits.

So, hundreds can be filled by 5 ways.

So, required number of ways in which three digit numbers can be formed from the given digits is $5\times5\times5=125 \ ways$

1(b) A three digit number is to be formed from given 5 digits 1,2,3,4,5.

Now, there are 5 ways to fill ones place.

Since, repetition is not allowed , so tens place can be filled by remaining 4 digits.

So, tens place can be filled in 4 ways.

Similarly, to fill hundreds place, we have 3 digits remaining.

So, hundreds can be filled by 3 ways.

So, required number of ways in which three digit numbers can be formed from the given digits is 5×4×3=60

2 Since repetition is not allowed

and the string consist of only four letters out of letters 'C' 'O' 'M' 'P' 'U' 'T' 'E' 'R' (all letters all unique)

So, first letter have 8 ways

and second letter of string have 7 ways (because one letter is already occupied)

third letter of string have 6 ways

and fourth letter of string have 5 ways

So, Total numbers of strings of length 4 can be formed using the letters COMPUTER = $8\times 7\times 6\times5=1680$

(3) Number of ways selecting "m" things out of "n" things = $^nC_r$

So, total number of ways of selecting 10 regions out of 17 = $^{17}C_{10}=$ 19448 ways

(4)

$\text{The expansion is given by the following formula:} \\(a+b)^n=∑(^nC_k)a^{n−k}b^k, \\where\ (^nC_k)=\dfrac{n!}{(n−k)!k!}\ and \ n!=1⋅2⋅…⋅n.\\ We\ have\ that\ \ a=2x, b=−y, n=3.\\ Therefore, (2x−y)^3=∑_{k=0}^3(^3C_k)(2x)^{3−k}(−y)^k.$

Now, calculate the product for every value of k from 0 to 3.

Upon Solving we get,

$(2x-y)^3=8x^3-12x^2y+6xy^2-y^3$

(5)(a)

$\text{The expansion is given by the following formula:} \\(a+b)^n=∑(^nC_k)a^{n−k}b^k, \\where\ (^nC_k)=\dfrac{n!}{(n−k)!k!}\ and \ n!=1⋅2⋅…⋅n.\\ We\ have\ that\ \ a=x, b=−2y, n=7.\\ Therefore, (x−2y)^7=∑_{k=0}^7(^7C_k)(x)^{7−k}(−2y)^k.$

On putting k= 3 in above formula and upon comparing we get

$(^7C_3)(x)^{7−3}(−2y)^3=-280x^4y^3$

So, Coefficient of $x^4y^3$ in expansion of $(x-2y)^7$ is equal to -280

(5)(b)

$\text{The expansion is given by the following formula:} \\(a+b)^n=∑(^nC_k)a^{n−k}b^k, \\where\ (^nC_k)=\dfrac{n!}{(n−k)!k!}\ and \ n!=1⋅2⋅…⋅n.\\ We\ have\ that\ \ a=2x, b=y, n=12.\\ Therefore, (2x+y)^{12}=∑_{k=0}^{12}(^{12}C_k)(2x)^{12−k}(y)^k.$

On putting k= 6 in above formula and upon comparing we get

$^{12}C_6(2x)^6(y)^6=59136 x^6y^6$

So, Coefficient of $x^6y^6$ in expansion of $(2x+y)^{12}$ is equal to 59136

(6)

$\text{The expansion is given by the following formula:} \\(a+b)^n=∑(^nC_k)a^{n−k}b^k, \\where\ (^nC_k)=\dfrac{n!}{(n−k)!k!}\ and \ n!=1⋅2⋅…⋅n.\\ We\ have\ that\ \ a=2x, b=y, n=12.\\ Therefore, (2x+y)^{12}=∑_{k=0}^{12}(^{12}C_k)(2x)^{12−k}(y)^k.$

For 9th term ,put k=8{since k is starting from 0 . So 9th term will be k=8}

$^{12}C_8(2x)^4(y)^8=7920x^4y^8$

Hence 9th term in the expansion of $(2x+y)^{12}$ is $7920x^4y^8$