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The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}
- a)400
- b)420
- c)800
- d)840
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The total number of injections (one-one into mappings) from {a1, a2, a...
So total = 7 × 6 × 5 × 4 = 840 one-one into functions
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The total number of injections (one-one into mappings) from {a1, a2, a...
Total number of injections from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7}
To find the total number of injections from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7}, we need to find the number of one-one mappings from set A to set B.
Method 1: Using the formula
The number of one-one mappings from set A to set B is given by the formula:
n!/(n-r)!
where n is the number of elements in set B and r is the number of elements in set A. In this case, n = 7 and r = 4. Substituting the values, we get:
7!/(7-4)! = 7!/3! = 7 x 6 x 5 x 4 = 840
Therefore, the total number of injections from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7} is 840.
Method 2: Using the multiplication principle
We can also find the number of injections using the multiplication principle. For each element in set A, we have 7 choices in set B. Since the mappings are one-one, we cannot repeat the choices. Therefore, the total number of injections is:
7 x 6 x 5 x 4 = 840
Therefore, both methods give the same answer of 840.
To find the total number of injections from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7}, we need to find the number of one-one mappings from set A to set B.
Method 1: Using the formula
The number of one-one mappings from set A to set B is given by the formula:
n!/(n-r)!
where n is the number of elements in set B and r is the number of elements in set A. In this case, n = 7 and r = 4. Substituting the values, we get:
7!/(7-4)! = 7!/3! = 7 x 6 x 5 x 4 = 840
Therefore, the total number of injections from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7} is 840.
Method 2: Using the multiplication principle
We can also find the number of injections using the multiplication principle. For each element in set A, we have 7 choices in set B. Since the mappings are one-one, we cannot repeat the choices. Therefore, the total number of injections is:
7 x 6 x 5 x 4 = 840
Therefore, both methods give the same answer of 840.
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The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}a)400b)420c)800d)840Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}a)400b)420c)800d)840Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}a)400b)420c)800d)840Correct answer is option 'D'. Can you explain this answer?.
The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}a)400b)420c)800d)840Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}a)400b)420c)800d)840Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of injections (one-one into mappings) from {a1, a2, a3 , a4} to {b1,b2 ,b3 ,b4 ,b5 ,b6 , b7 is}a)400b)420c)800d)840Correct answer is option 'D'. Can you explain this answer?.
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