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gtnhenbr [62]
3 years ago
8

Evaluate the expressions. 8!-6!, 7!-5!, and 12!/8! BRAINLIST 5 min left plz hurry

Mathematics
1 answer:
nordsb [41]3 years ago
4 0

Answer:

The evaluated expressions are 8!-6!=39600 ,   7!-5!=4920  and \frac{12!}{8!}=11880

Step-by-step explanation:

The factorial of a positive integer n is the product of all positive integers less than or equal to n:

n!=n\times (n-1)\times (n-2)\times (n-3)\times ...\times 3\times 2\times 1

The factorial of n is denoted by n!

Now to find  the expressions 8!-6!, 7!-5!, and 12!/8! :

Given expression is 8!-6!:

8!-6!=[8\times (8-1)\times (8-2)\times (8-3)\times (8-4)\times (8-5)\times (8-6)\times (8-7)]- [6\times (6-1)\times (6-2)\times (6-3)\times (6-4)\times (6-5)]  (By using factorial of n formula)

=[8\times 7\times 6\times 5\times 4\times 3\times 2\times 1]- [6\times 5\times 4\times 3\times 2\times 1]

=[6\times 5\times 4\times 3\times 2\times 1][(8\times 7)-1]

=[720][56-1]

=[720][55]

=39600

Therefore  8!-6!=39600

Given expression is 7!-5! :

7!-5!=[7\times (7-1)\times (7-2)\times (7-3)\times (7-4)\times (7-5)\times (7-6)]-[5\times (5-1)\times (5-2)\times (5-3)\times (5-4)]  (By using factorial of n formula)

=[7\times 6\times 5\times 4\times 3\times 2\times 1]- [5\times 4\times 3\times 2\times 1]

=[5\times 4\times 3\times 2\times 1][(7\times 6)-1]

=[120][42-1]

=[120][41]

=4920

Therefore  7!-5!=4920

Given expression is \frac{12!}{8!}

\frac{12!}{8!}=\frac{12\times (12-1)\times (12-2)\times (12-3)\times (12-4)\times (12-5)\times (12-6)\times(12-7)\times (12-8)\times (12-9)\times (12-10) \times(12-11)}{8\times (8-1)\times (8-2)\times (8-3)\times (8-4)\times(8-5)\times (8-6)\times (8-7)}  (By using factorial of n formula)

 =\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times4\times 3\times 2\times 1}{8\times 7\times 6\times 5\times4\times 3\times 2\times 1}

 =12\times 11\times 10\times 9

=11880

Therefore  \frac{12!}{8!}=11880

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The  sequence  is:

10, 30, 50, 70, 90.....................

Step-by-step explanation:

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First term (a) = 10

Common difference (d) = ?

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or, \frac{n}{2} [{2a+(n-1)d}] = 250

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3 years ago
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