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The number of ways in which 33 identical pens can be distributed among three boys such that each of them receives an odd number of pens is
- a)136
- b)156
- c)150
- d)105
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The number of ways in which 33 identical pens can be distributed among...
Let 2a+1, 2b+1, 2c+1 be the number of pens received by each of the three boys.
2a+1+2b+1+2c+1=33 [0 ≤ a, b, c ≤ 15]
2a+2b+2c=30
a+b+c=15
Required number of ways = 15 + 3 - 1C3 - 1
= 17C2
= 136
A is the correct answer.
2a+1+2b+1+2c+1=33 [0 ≤ a, b, c ≤ 15]
2a+2b+2c=30
a+b+c=15
Required number of ways = 15 + 3 - 1C3 - 1
= 17C2
= 136
A is the correct answer.
Most Upvoted Answer
The number of ways in which 33 identical pens can be distributed among...
Solution:
Given that there are 33 identical pens to be distributed among three boys such that each of them receives an odd number of pens.
Let the number of pens received by the three boys be x, y and z respectively.
Therefore, x + y + z = 33 (as the total number of pens is 33)
Also, x, y and z are odd numbers.
So, let x = 2a + 1, y = 2b + 1 and z = 2c + 1, where a, b and c are integers.
Substituting the values of x, y and z in the equation x + y + z = 33, we get
2a + 2b + 2c + 3 = 33
2(a + b + c) = 30
a + b + c = 15
Now, we need to find the number of solutions for the above equation.
We can use the formula for distributing n identical objects among r recipients, such that each recipient receives at least one object, which is:
Number of Solutions = (n-1)C(r-1)
Here, n = 15 and r = 3
Number of Solutions = (15-1)C(3-1)
= 14C2
= 91
Now, we need to find the number of ways in which we can assign the values of a, b and c to x, y and z.
For this, we can use another formula, which is:
Number of Solutions = (n+r-1)C(r-1)
Here, n = 3 (as there are 3 boys) and r = 3 (as we need to assign values to x, y and z)
Number of Solutions = (3+3-1)C(3-1)
= 5C2
= 10
Therefore, the total number of ways in which 33 identical pens can be distributed among three boys such that each of them receives an odd number of pens is:
Total number of ways = Number of Solutions for the equation a + b + c = 15 x Number of Solutions for assigning values to x, y and z
= 91 x 10
= 910
Hence, the correct answer is option A) 136.
Given that there are 33 identical pens to be distributed among three boys such that each of them receives an odd number of pens.
Let the number of pens received by the three boys be x, y and z respectively.
Therefore, x + y + z = 33 (as the total number of pens is 33)
Also, x, y and z are odd numbers.
So, let x = 2a + 1, y = 2b + 1 and z = 2c + 1, where a, b and c are integers.
Substituting the values of x, y and z in the equation x + y + z = 33, we get
2a + 2b + 2c + 3 = 33
2(a + b + c) = 30
a + b + c = 15
Now, we need to find the number of solutions for the above equation.
We can use the formula for distributing n identical objects among r recipients, such that each recipient receives at least one object, which is:
Number of Solutions = (n-1)C(r-1)
Here, n = 15 and r = 3
Number of Solutions = (15-1)C(3-1)
= 14C2
= 91
Now, we need to find the number of ways in which we can assign the values of a, b and c to x, y and z.
For this, we can use another formula, which is:
Number of Solutions = (n+r-1)C(r-1)
Here, n = 3 (as there are 3 boys) and r = 3 (as we need to assign values to x, y and z)
Number of Solutions = (3+3-1)C(3-1)
= 5C2
= 10
Therefore, the total number of ways in which 33 identical pens can be distributed among three boys such that each of them receives an odd number of pens is:
Total number of ways = Number of Solutions for the equation a + b + c = 15 x Number of Solutions for assigning values to x, y and z
= 91 x 10
= 910
Hence, the correct answer is option A) 136.
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The number of ways in which 33 identical pens can be distributed among three boys such that each of them receives an odd number of pens isa)136b)156c)150d)105Correct answer is option 'A'. Can you explain this answer?
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