Question:
A box contains 24 identical balls of which one ball is black and the remaining balls are green. Three balls are taken simultaneously and randomly. The number of ways of getting only green balls, is
A box contains 24 identical balls of which one ball is black and the remaining balls are green. Three balls are taken simultaneously and randomly. The number of ways of getting only green balls, is
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When unwanted items exist, select only from required group directly.
Updated On: Apr 24, 2026
- $1765$
- $1764$
- $1763$
- $1771$
- $1864$
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
• Use combinations: ${}^nC_r = \dfrac{n!}{r!(n-r)!}$
• Only green balls must be selected
Step 1: Identify total green balls
\[ 24 - 1 = 23 \]
Step 2: Select 3 green balls
\[ {}^{23}C_3 = \frac{23 \times 22 \times 21}{6} = 1771 \] Final Conclusion:
\[ = 1771 \]
• Use combinations: ${}^nC_r = \dfrac{n!}{r!(n-r)!}$
• Only green balls must be selected
Step 1: Identify total green balls
\[ 24 - 1 = 23 \]
Step 2: Select 3 green balls
\[ {}^{23}C_3 = \frac{23 \times 22 \times 21}{6} = 1771 \] Final Conclusion:
\[ = 1771 \]
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