Question

The number of ways six people can be placed in a line for a photo can be determined using the expression 6!. What is the value of 6!? Two of the six people are given responsibilities during the photo shoot. One person holds a sign and the other person points to the sign. The expression represents the number of ways the two people can be chosen from the group of six. In how many ways can this happen? In the next photo, three of the people are asked to sit in front of the other people. The expression represents the number of ways the group can be chosen. In how many ways can the group be chosen?

Answer 1

Answer:

Step-by-step explanation:

Using n!=n(n-1)(n-2)----1

Value of 6! is   $6\cdot5\cdot 4\cdot 3\cdot2\cdot 1=720$

Now, In second case we have to choose two people out of 6 for the photo shoot it will be

Here, n that is total possibilities is 6

And the specific chosen possibility that is r is 2

By using $^{n}C_r=\frac{n!}{r!\cdot(n-r)!}$

On substituting the values we get:

$^{6}C_2=\frac{6!}{2!\cdot(6-2)!}$

$\Rightarrow ^{6}C_2=\frac{6!}{2!\cdot(4)!}$

$\Rightarrow ^{6}C_2=\frac{6\cdot5 \cdot4!}{2!\cdot(4)!}$

Cancel the common terms we  we get:

$\Rightarrow ^{6}C_2=\frac{6\cdot5}{2\cdot 1}=15$

Now, third case is 3 students out of 6 are to be seated in front row it will be

$^{6}C_3=\frac{6!}{3!\cdot(6-3)!}$

$\Rightarrow ^{6}C_3=\frac{6!}{3!\cdot(3)!}$

$\Rightarrow ^{6}C_3=\frac{6\cdot5 \cdot4\cdot 3!}{3\cdot2 \cdot1\cdot(3)!}$

Cancel out the common factors we get;

$\Rightarrow ^{6}C_3=\frac{6\cdot5 \cdot4}{3\cdot2 \cdot1}$$\Rightarrow ^{6}C_3=20$

Answer 2

The value of 6! is 720 and the number of ways two of six people are given responsibilities is 15

Factorials are calculated using:

• $n! = n \times (n -1) \times (n -2) \times ....... \times 2 \times 1$

To calculate 6!, we make use of:

$6! = 6 \times 5 \times 4 \times 3\times 2 \times 1$

$6! = 720$

So, the value of 6! is 720

Next, we calculate the number of ways two of six people are given responsibilities using:

$^nC_r = \frac{n!}{(n - r)!r!}$

This gives

$^6C_2 = \frac{6!}{(6 - 2)!2!}$

$^6C_2 = \frac{6!}{4!2!}$

Expand 6! and 2!

$^6C_2 = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1}$

$^6C_2 = \frac{6 \times 5 }{2 \times 1}$

$^6C_2 = \frac{30}{2}$

$^6C_2 = 15$

Hence, the number of ways two of six people are given responsibilities is 15

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