Ex7.1, 6 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other? A signal can have only 2 flags The required number of signals = 5 × 4 = 20
Show MoreThat was with 4flags ,but since here its mentioned at least 4, so , 5 or 6 flags can also be used
No. of signals with 4 flags = 6P4=360
No. of signals with 5 flags = 6P5=6!=720
No. of signals with 6 flags = 6P6=720
Total no. of signals = 720+720+360=1800
Complete step by step answer:
Now we have 4 flags of different colours. Signals can be generated by choosing 2 flags.
Hence we will first select 2 flags out of 4 flags.
Now we know that the number of ways of selecting r objects from n objects is $^{n}{{C}_{r}}$ .
Where $^{n}{{C}_{r}}=\dfrac{n!}{[n-r]!r!}$ and $a!=a\times [a-1]\times [a-2]\times ....\times [2]\times 1$
Hence number of ways of selecting 2 flags from 4 flags is given by $^{4}{{C}_{2}}$
$^{4}{{C}_{2}}=\dfrac{4!}{[4-2]!2!}=\dfrac{4\times 3\times 2}{2\times 2}=2\times 3=6$
Hence the number of ways of choosing 2 flags out of 4 flags is 6 …………… [1]
Now we will arrange this 2 flags
We know that number of ways to arrange n objects is n!
Hence we can arrange these two flags in 2! = 2 ways ……………… [2]
Now from equation [1] and equation [2] we get that total number of ways = 6 × 2 = 12.
Hence we have the total number of signals possible is 12.
Note:
We can also think of this problem in a different manner. Let us say we have 4 flags named A, B, C and D
Now first let us say we have flag A above, then we can have B, C, D below hence we have 3 choices.
Similarly if we have a B flag above then also we have 3 choices.
Same for C and D we will have 3 choices for each.
Hence the total possible signal is 3 + 3 + 3 + 3 = 12.
Hence we have a total number of possible signals is 12.