There are 8 different pens and 4 different pencils. In how many ways can 4 pens and 2 pencils be arranged?
1
Answers
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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.