GO Classes Test Series 2024 | Discrete Mathematics | Test 4 | Question: 25

Answer 1

Score of bit string must be $10$ means we must have ten $1$’s in the bit string. Bit string must start and end with $1,$ so, directly put $1$’s in the left most and right most position, so, we have 8 more $1$’s to put between them. So, basically the question is now that we have eight $1$’s and no two zeroes should be consecutive.

Now we have following cases :

No $0$’s OR exactly one $0$ OR exactly two $0$’s Or $\dots$ exactly nine $0\text{’s}.$

$^{9}\text{C}_{0} + ^{9}\text{C}_{1} + ^{9}\text{C}_{2} + \dots + ^{9}\text{C}_{9} = 2^9 = 512.$

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$\textbf{Alternative Method:}$ We put of $10$ ones by keeping the gap between them, shown in the following picture :



Each square can be either left empty or we can replace it by $a\; 0.$
At each square, we can either leave it blank or replace it by $a \;0.$

Clearly, there are $2^{(9)} = 512$ ways to do this.

Answer 2

here score is what the no of ones in the string

we are given that

string must contains 10’s 1

and

the string must start  and end with 1

and

2 0s cant be consecutive so we have to fill zeros in the gap

1 _1 _1_1 _1 _1 _1 _1 _ 1_1

so here we have 10’s 1 and 9 gap

so we can fill 9 gaps with 0 zero,1zeros,2 zeros,     9 zeros

filling  9 gaps with no zero=9C0

filling 9 gaps with one zero=9c1

filling 9 gaps with 2 zero=9C2

filling 9 gaps with 3 zero=9C3

filling 9 gaps with 4 zero=9C4

filling 9 gaps with 5 zero=9C5

filling 9 gaps with 6 zero=9C6

filling 9 gaps with 7 zero=9C7

filling 9 gaps with 8 zero=9C8

filling 9 gaps with 9 zero=9C9

so total combinations possible is =9C0+9C1+9C2+9C3+9C4+9C5+9C6+9C7+9C8+9C9

                                                     =2^(9)

                                                      =512