# Question: When n is large n can be approximated by means

When n is large, n! can be approximated by means of the expression

Called Stirling’s formula, where e is the base of natural logarithms. (A derivation of this formula may be found in the book by W. Feller cited among the references at the end of this chapter.)

(a) Use Stirling’s formula to obtain approximations for 10! and 12!, and find the percentage errors of these approximations by comparing them with the exact values given in Table VII.

(b) Use Stirling’s formula to obtain an approximation for the number of 13-card bridge hands that can be dealt with an ordinary deck of 52 playing cards.

Called Stirling’s formula, where e is the base of natural logarithms. (A derivation of this formula may be found in the book by W. Feller cited among the references at the end of this chapter.)

(a) Use Stirling’s formula to obtain approximations for 10! and 12!, and find the percentage errors of these approximations by comparing them with the exact values given in Table VII.

(b) Use Stirling’s formula to obtain an approximation for the number of 13-card bridge hands that can be dealt with an ordinary deck of 52 playing cards.

## Relevant Questions

Using Stirling’s formula (see Exercise 1.6) to approximate 2n! and n!, show that It is known from experience that in a certain industry 60 percent of all labor– management disputes are over wages, 15 percent are over working conditions, And 25 percent are over fringe issues. Also, 45 percent of the ...An art dealer receives a shipment of five old paintings from abroad, And, on the basis of past experience, she feels that the probabilities are, respectively, 0.76, 0.09, 0.02, 0.01, 0.02, And 0.10 that 0, 1, 2, 3, 4, or all ...Show that the postulates of probability are satisfied by conditional probabilities. In other words, show that if P(B) ≠ 0, then (a) P(A| B) ≥ 0; (b) P(B| B) = 1; (c) P(A1 ∪ A2 .. ∪ . | B) = P(A1| B) + P(A2| B) + · ...If A1, A2,…, An are independent events, prove that P(A1 ∪ A2,···, An) = 1-{1- P(A1) } · {1- P(A2) } … {1- P(A ∩) }Post your question