CA Foundation Exam > CA Foundation Questions > The value of 12C4 + 12C3 isa)715b)710c)716d)n...
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The value of 12C4 + 12C3 is
- a)715
- b)710
- c)716
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The value of 12C4 + 12C3 isa)715b)710c)716d)none of theseCorrect answe...
Solution:
We know that nCr = nC(n-r)
So, 12C4 = 12C(12-4) = 12C8
Also, 12C3 = 12C(12-3) = 12C9
Now, we can use the formula for combination:
nCr = n!/r!(n-r)!
Using this formula, we can calculate:
12C8 = 12!/8!4! = (12*11*10*9)/(4*3*2*1) = 495
12C9 = 12!/9!3! = (12*11*10)/(3*2*1) = 220
Therefore, the value of 12C4 * 12C3 = 495 * 220 = 108900
Hence, option A (715) is incorrect and the correct answer is option A (108900).
We know that nCr = nC(n-r)
So, 12C4 = 12C(12-4) = 12C8
Also, 12C3 = 12C(12-3) = 12C9
Now, we can use the formula for combination:
nCr = n!/r!(n-r)!
Using this formula, we can calculate:
12C8 = 12!/8!4! = (12*11*10*9)/(4*3*2*1) = 495
12C9 = 12!/9!3! = (12*11*10)/(3*2*1) = 220
Therefore, the value of 12C4 * 12C3 = 495 * 220 = 108900
Hence, option A (715) is incorrect and the correct answer is option A (108900).
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Community Answer
The value of 12C4 + 12C3 isa)715b)710c)716d)none of theseCorrect answe...
NCr = n! / r! (n-r)!
12C4 = (12×11×10×9) / (4×3×2×1) = 495
12C3 = (12×11×10) / (3×2×1) = 220
12C4 + 12C3 = 495 + 220 = 715
Hence the correct answer is option A) 715.
12C4 = (12×11×10×9) / (4×3×2×1) = 495
12C3 = (12×11×10) / (3×2×1) = 220
12C4 + 12C3 = 495 + 220 = 715
Hence the correct answer is option A) 715.
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