Question

How many combinations without repetition are possible if n = 4 and r = 3?

Answer 1

The possible combinations without repetition are 4.

Given

n = 4

r = 3

To Find

Possible Combinations without repetition

Solution

The numerical can be solved by following the given steps -

Here, it is given that

n = 4

r = 3

For combinations without repetition, we use the formula

$C_{(n,r)} = \frac{n!}{r! (n-r)!}$

For the given question, the possible combination is given as

$C_{(n,r)} = \frac{4!}{3! (4-3)!} \\\\C_{(4,3)} = \frac{4*3*2*1}{3*2*1 (1)} \\\\C_{(4,3)} = 4$

Therefore, the possible combinations without repetition are 4.

Answer 2

Several combinations without repetition are possible 4.

Given:

n = 4 and r = 3.

To Find:

Several combinations without repetition are possible.

Solution:

To find the number of combinations without repetition we will follow the following steps:

As we know,

The formula used for finding combinations without repetition is

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

Here, n is the total number of elements while r is the number of elements taken for combination or permutation.

$\frac{n!}{r! (n- r)!} = \frac{4!}{3! (4 - 3)!} = \frac{4 \times 3×2×1}{3×2×1 \times 1} = 4$

Henceforth, The number of combinations possible without repetition is 4.

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