Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.2

Question 1. Find x if \( \frac{1}{6 !} + \frac{1}{7 !} = \frac{x}{8 !} \)
Answer: To solve this, we can write the larger factorials in terms of the smallest one, which is \( 6! \).
So, \( \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!} \)
We know that \( 7! = 7 \times 6! \) and \( 8! = 8 \times 7 \times 6! \).
Substitute these into the equation:
\( \frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8 \times 7 \times 6!} \)
Next, we multiply the entire equation by \( 6! \) to remove the denominators involving factorials.
\( 1 + \frac{1}{7} = \frac{x}{8 \times 7} \)
Now, combine the terms on the left side:
\( \frac{7+1}{7} = \frac{x}{56} \)
\( \frac{8}{7} = \frac{x}{56} \)
To find x, multiply both sides by 56:
\( x = \frac{8}{7} \times 56 \)
\( x = 8 \times 8 \)
\( x = 64 \)
In simple words: To find x, we first rewrite the numbers with factorials so they all have a common part. Then we cancel out that common part and solve the remaining simple equation to get the value of x.

🎯 Exam Tip: When dealing with factorials in an equation, always try to express the larger factorials in terms of the smallest one to simplify cancellation and make calculations easier.

Question 2. Evaluate \( \frac{n !}{r !(n-r) !} \) when n = 5 and r = 2.
Answer: We are given the formula \( \frac{n!}{r!(n-r)!} \), which is used for combinations. We need to find its value when \( n = 5 \) and \( r = 2 \).
First, substitute the values of n and r into the formula:
\( \frac{5!}{2!(5-2)!} \)
Now, simplify the term in the parenthesis:
\( \frac{5!}{2!3!} \)
Next, expand the factorials. Remember that \( n! = n \times (n-1) \times \dots \times 1 \).
\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
\( 2! = 2 \times 1 = 2 \)
\( 3! = 3 \times 2 \times 1 = 6 \)
Substitute these expanded values back into the expression:
\( \frac{120}{2 \times 6} \)
\( \frac{120}{12} \)
\( = 10 \)
Thus, the value of the expression is 10.
In simple words: We put the given numbers (n=5, r=2) into the formula. We then work out each factorial part (like 5! means 5x4x3x2x1). Finally, we divide the numbers to get the answer.

🎯 Exam Tip: Always remember the definition of a factorial and carefully calculate each part. For this specific formula, it's often easier to expand the largest factorial up to the next largest one in the denominator and then cancel.

Question 3. If \( (n + 2)! = 60[(n - 1)!] \), find n.
Answer: We need to find the value of n in the equation \( (n + 2)! = 60(n - 1)! \).
We can expand the larger factorial, \( (n+2)! \), until we reach \( (n-1)! \).
\( (n+2)! = (n+2) \times (n+1) \times n \times (n-1)! \)
Now substitute this into the given equation:
\( (n+2)(n+1)n(n-1)! = 60(n-1)! \)
We can cancel \( (n-1)! \) from both sides of the equation, as long as \( (n-1)! \) is not zero, which means \( n-1 \ge 0 \), so \( n \ge 1 \).
\( (n+2)(n+1)n = 60 \)
We are looking for three consecutive integers whose product is 60. We can try to factor 60 to find these integers.
\( 60 = 5 \times 12 \)
\( 60 = 5 \times 4 \times 3 \)
So, we have \( (n+2)(n+1)n = 5 \times 4 \times 3 \)
By comparing the terms, we can see that \( n = 3 \).
If \( n=3 \), then \( n+1 = 4 \) and \( n+2 = 5 \). This matches the product.
Thus, \( n = 3 \).
In simple words: We rewrite the bigger factorial part to match the smaller one. Then we cancel out the common part. This leaves us with a product of three numbers that equals 60. By figuring out which three consecutive numbers multiply to 60, we find n.

🎯 Exam Tip: Always expand the larger factorial down to match the smaller factorial in the expression. This simplifies the equation significantly and often leads to finding consecutive integers.

Question 4. How many five digits telephone numbers can be constructed using the digits 0 to 9 If each number starts with 67 with no digit appears more than once?
Answer: We need to form a five-digit telephone number. The numbers must start with 67, and no digit can be repeated.
The total digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (a total of 10 digits).
The first two digits are fixed as 6 and 7. This means we have used two digits.
Since no digit can appear more than once, digits 6 and 7 cannot be used again for the remaining three places.
The number of remaining digits is \( 10 - 2 = 8 \) digits (0, 1, 2, 3, 4, 5, 8, 9).
We need to fill the remaining three places (the third, fourth, and fifth digits).
For the third digit: We have 8 choices from the remaining digits.
For the fourth digit: Since one digit is used for the third place, we have \( 8 - 1 = 7 \) choices left.
For the fifth digit: Since two digits are used for the third and fourth places, we have \( 7 - 1 = 6 \) choices left.
By the multiplication principle, the total number of telephone numbers is the product of the number of choices for each place:
Total numbers = (choices for 3rd digit) \( \times \) (choices for 4th digit) \( \times \) (choices for 5th digit)
Total numbers = \( 8 \times 7 \times 6 \)
Total numbers = \( 336 \)
So, 336 different five-digit telephone numbers can be constructed.
In simple words: We have a five-digit phone number that must start with 67. We can't use numbers more than once. Since 6 and 7 are used, we have 8 digits left for the next spot, then 7 for the one after that, and 6 for the last spot. We multiply these choices together to get the total number of unique phone numbers.

🎯 Exam Tip: For problems with 'no repetition' or 'distinct digits', remember to decrease the number of available choices for each subsequent position you fill. Always account for any pre-fixed digits first.

Question 5. How many numbers lesser than 1000 can be formed using the digits 5, 6, 7, 8, and 9 if no digit is repeated?
Answer: We need to form numbers less than 1000 using the digits 5, 6, 7, 8, and 9. This means we can form one-digit, two-digit, or three-digit numbers. Also, no digit can be repeated.
There are a total of 5 distinct digits available.

1. **One-digit numbers:** Any of the 5 digits (5, 6, 7, 8, 9) can be a one-digit number. Number of one-digit numbers = 5.

2. **Two-digit numbers:** For the tens place, we have 5 choices (any of 5, 6, 7, 8, 9). For the units place, since no digit can be repeated, we have \( 5 - 1 = 4 \) choices remaining. Number of two-digit numbers = \( 5 \times 4 = 20 \).

3. **Three-digit numbers:** For the hundreds place, we have 5 choices. For the tens place, we have \( 5 - 1 = 4 \) choices remaining. For the units place, we have \( 4 - 1 = 3 \) choices remaining. Number of three-digit numbers = \( 5 \times 4 \times 3 = 60 \).

To find the total number of numbers less than 1000, we add the numbers from each case:
Total numbers = (one-digit numbers) + (two-digit numbers) + (three-digit numbers)
Total numbers = \( 5 + 20 + 60 = 85 \).
So, 85 numbers less than 1000 can be formed.
In simple words: We can make one-digit, two-digit, or three-digit numbers from the given digits without repeating any. We calculate how many of each type we can make (5 single-digit, 20 two-digit, 60 three-digit). Then, we add all these counts together to get the total.

🎯 Exam Tip: When forming numbers under a certain limit (like 1000), break the problem into cases based on the number of digits. Always remember to reduce the number of choices for each position when repetition is not allowed.