Question

Find the number of permutations of the first 9 letters of the alphabet, taking 4 letters at a time.

Answer 1

Solution-

First 9 letters of alphabet are- A,B,C,D,E,F,G,H,I

Total number of ways of selection of 4 letters from 9 alphabets = 9C4

=9!÷((4!).(9-4)!) = 9!÷(4!×5!) = (9×8×7×6)÷(24) = 126

The number of ways of arranging these 4 numbers = 4! = 24

∴ Total number of possible permutations = 9C4×4! = 126×24 = 3024

∴ option number 2 is correct.

Answer 2

Answer:

3024

Step-by-step explanation:

The combinations to find the number of permutations of the first 9 letters of the alphabets, taking 4 letters at one time, are given by:

$_nC_k$

$= \frac{_nP_k}{k!}$

$(\frac{n!}{k!(n-k)!} )$

Here in this case,

n = 9 (the total number of letters);and

k = 4 (the numbers of letters taken at one time)

$_9C_4 = (\frac{9!}{4!.5!})$

$9 . 8 . 7 . 6 . 5 .\frac{4!}{4!.5!}$

$9.8.7.6. \frac{5!}{5!}$

$9.8.7.6 = 3024$

Therefore, the number of permutations of the first 9 letters of the alphabets is 3024.