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Permutations and Combinations

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Question : 24 of 56

Marks: +1, -0

Evaluate

:{}^{47}C_{4} + \sum\limits_{j=0}^{3} (50-j) C_{3}

249900
24990
249000
None

Solution:

Concept:

Use binomial coefficient formula and summation of combinations.

Explanation:

{}^{47}C_4 = \frac{47 \cdot 46 \cdot 45 \cdot 44}{4 \cdot 3 \cdot 2 \cdot 1} = 178365

.

The sum

\sum_{j=0}^{3} {}^{(50-j)}C_3 = {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3

.

{}^{50}C_3 = \frac{50 \cdot 49 \cdot 48}{6} = 19600

{}^{49}C_3 = \frac{49 \cdot 48 \cdot 47}{6} = 18424

{}^{48}C_3 = \frac{48 \cdot 47 \cdot 46}{6} = 17296

{}^{47}C_3 = \frac{47 \cdot 46 \cdot 45}{6} = 16215

Sum of these four = 19600 + 18424 + 17296 + 16215 = 71535.

Total =

178365 + 71535 = 249900

.

Answer:

249900

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