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Examples from Permutations
Data Science and A.I. Lecture Series
By Bindeshwar Singh Kushwaha, PostNetwork Academy
Example 1
How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution:
- Total digits: 9
- Required 4-digit numbers:
\[
P(9, 4) = \frac{9!}{(9-4)!} = 9 \times 8 \times 7 \times 6 = 3024
\]
Example 2
How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5 if the repetition of the digits is not allowed?
Solution:
- Total 3-digit numbers:
\[
P(6, 3)
\] - Numbers with 0 in the hundreds place:
\[
P(5, 2)
\] - Required numbers:
\[
P(6, 3) – P(5, 2) = 120 – 20 = 100
\]
Example 3
Find the value of \( n \) such that \( P(n, 5) = 42 \times P(n, 3) \).
Solution:
- Expand:
\[
n(n-1)(n-2)(n-3)(n-4) = 42 \times n(n-1)(n-2)
\] - Simplify:
\[
n^2 – 7n – 30 = 0
\] - Solve:
\[
(n-10)(n+3) = 0 \Rightarrow n = 10 \quad (\text{as } n > 0)
\]
Example 4
Find \( r \) if \( P(5, r) = 6 \times P(5, r-1) \).
Solution:
- Expand:
\[
\frac{5!}{(5-r)!} = 6 \times \frac{5!}{(5-(r-1))!}
\] - Simplify:
\[
(6-r)(5-r) = 6
\] - Solve:
\[
r^2 – 11r + 24 = 0 \Rightarrow (r-8)(r-3) = 0
\] - Values of \( r \): \( r = 8 \) or \( r = 3 \).
Example 5
Find the number of different 8-letter arrangements of DAUGHTER where:
- (i) All vowels occur together.
- (ii) All vowels do not occur together.
Solution:
- (i) Treat vowels (A, U, E) as a single unit:
\[
6! \times 3! = 4320
\] - (ii) Total arrangements:
\[
8! – 6! \times 3! = 36000
\]
Example 6
In how many ways can 4 red, 3 yellow, and 2 green discs be arranged in a row if discs of the same color are indistinguishable?
Solution:
- Total discs: 9
- Arrangements:
\[
\frac{9!}{4! \times 3! \times 2!} = 1260
\]
PDF Presentation
MoreExamplesonPermutationsVideo
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