Divinefox528

Resources from the previous (2012-2014) unit plans are still available on the secondary curriculum website; however the alignment varies from the updated scope & sequence.

The standards for mathematical practice describe varieties of expertise that mathematics educators at all levels should seek to develop in their students. These practices rest on important “processes and proficiencies” with longstanding importance in mathematics education. For more detailed information about unpacking the content standards, unpacking a task, math routines and rituals, maintenance activities and more, please refer to the Grade Level Overview. In this unit, students will.

1.1

1.

Soln:

A man can enter the stadium in 4 ways. Again the man can leave the stadium in 9 ways.

So, total no.of ways with which a man enters and then leaves the stadium = 4 * 9 = 36ways.

2.

Soln:

There are 6 choices for a student to enter the hostel. There are 5 choices for a student to leave the hostel as different door is to be used.

So, total no.of ways = 6 * 5 = 30.

3.

Soln:

There are 7 choices for 1^st son, 6 choices for 2^nd son and 5 choices for 3^rd son.

Now, by the basic principle of counting, the total number of ways of choice = 7 * 6 * 5 = 210.

4.

Soln:

A man can go from city A to city B in 5 ways. As he has to return by a different road, so he can return from city B to city A in 4 ways.

So, total no.of ways by which a man can go from city A to city B and returns by a different road = 5 * 4 = 20 ways.

5.

Soln:

A person can go from city A to city B in 5 ways. Again, he can go from city B to city C in 4 ways. So, a person can go from city A to city C in 5 * 4 = 20ways. The person has to return from C to A without driving on the same road twice, So, he can return from city C to city B in 3 ways and from city B to city A in 4 ways.

Old Unit4 Agendasmrs. Colville's Math Classes

So, he can return from city C to city A in 3 * 4 = 12 ways.

So, Total no.of ways by which a person can go from city A to city C and return from city C to city A = 20 * 12 = 240 ways.

6.

Soln:

Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.

There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively.

So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120

Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360.

the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720.

The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720.

So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.

7.

Soln:

The numbers formed must be of three digits and less than 500, so the digit in the hundred’s place should be 1,2,3 or 4. So, there are 4 choices for the digit in the hundred’s place. There are 5 choice for the digit in the ten’s place. There are 4 choices for the digit in the unit’s place.

So, no of ways by which 3 digits numbers les than 500 can be formed = 4.5.4 = 80.

8.

Soln:

The numbers formed should be even. So, the digit in the unit’s place must be 2 or 4. So, the digit in unit’s place must be 2 or 4. So, for the digit in unit’s place, there are 2 choices. So, after fixing the digit in the unit’s place, remaining 4 figures can be arranged in P(4,4) ways.

Ie. $frac{{left( 4 right)!}}{{left( {4 - 4} right)!}}$ = $frac{{4!}}{{0!}}$ = $frac{{4{rm{*}}3{rm{*}}2{rm{*}}1}}{1}$ = 24 ways.

So, total no.of ways by which 5 even numbers can be formed = 2 * 24 = 48.

9.

Soln:

The numbers formed must be of 4 digits. The digit in the thousand’s place must always be 4. For this, there is only one choice. After that, n = 6 – 1 = 5, r = 4 – 1 = 3. Then remaining 5 figures can be placed in remaining 3 places in:

Or, P(5,3) ways = $frac{{5!}}{{left( {5 - 3} right)!}}$ = $frac{{5!}}{{2!}}$ = $frac{{5{rm{*}}4{rm{*}}3{rm{*}}2{rm{*}}1}}{{2{rm{*}}1}}$ = 60 ways.

So, Total no.of ways by which 4 digits numbers between 4,000 and 5,000 can be formed = 1 * 60 = 60.

10.

Soln:

For the three digits numbers, there are 5 ways to fill in the 1^st place, there are 4 ways to fill in the 2^nd place and there are 3 ways to fill in the 3^rd place. By the basic principle of counting, number of three digits numbers = 5 * 4 * 3 = 60.

Old Unit4 Agendasmrs. Colville's Math Classroom

Again, for three digit numbers which are divisible by 5, the number in the unit place must be 5. So, the unit place can be filled up in 1 way. After filling up the unit place 4 numbers are left. Ten’s place can be filled up in 4 ways and hundredths place can be filled up in 3 ways. Then by the basic principle of counting, no.of 3 digits numbers which are divisible by 5 = 1 * 4 * 3 = 12.