Computer Science, asked by monieoui, 6 months ago

There are 8 different pens and 4 different pencils. In how many ways can 4 pens and 2 pencils be arranged?

1​

Answers

Answered by itzBrainlymaster
1

please see the picture dear...

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Answered by sakshi15982
0

Answer:

Answer: The required number of ways is 11760.

Step-by-step explanation: We are given to find the number of ways in which a bag consisting of 5 pens and 6 pencils can be filled from a stock having 8 pens and 10 pencils.

Since the arrangements of the pens and pencils does not matter, so this is a problem of combination.

We have

The number of ways in which 5 pens can be selected from 8 pens is given by

^8C_5=\dfrac{8!}{5!(8-5)!}=\dfrac{8\times7\times6\times5!}{5!\times3\times2\times1}=56

8

C

5

=

5!(8−5)!

8!

=

5!×3×2×1

8×7×6×5!

=56

and

the number of ways in which 6 pencils can be selected from 10 pencils is given by

^{10}C_6=\dfrac{10!}{6!(10-6)!}=\dfrac{10\times9\times8\times7\times6!}{6!\times4\times3\times2\times1}=210.

10

C

6

=

6!(10−6)!

10!

=

6!×4×3×2×1

10×9×8×7×6!

=210.

Therefore, the number of ways in which a bag consisting of 5 pens and 6 pencils can be filled from a stock having 8 pens and 10 pencils is given by

n=^8C_5\times^{10}C_6=56\times210=11760.n=

8

C

5

×

10

C

6

=56×210=11760.

Thus, the required number of ways is 11760.

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