(I) How many TEN-BIT strings of 0\'s and 1\'s are there that begin or end with a

Question

(I) How many TEN-BIT strings of 0's and 1's are there that begin or end with a 1 (or both)? (J) In how many ways can a committee of five persons be selected from eight men and seven women such that at least one man is on the committee? (K) In how many ways can a committee of four persons be chosen from a group of ten men and seven women such that the committee does not have the same number of men on the committee as women (two each)? (L) The University of Tulsa phone numbers look like 631-xxxx where xxxx is the 4-digit extension. How many 4-digit extension numbers are there that have at least one 7 in the extension number? (M) A quiz has 5 questions. How many ways can (integer) points be assigned to the problems if the total points is 20 and each problem is worth at least two points each?

Explanation / Answer

Answer :

(I) Answer would be: 28

Because we know, that the bit starts with1 and ends with 1,that means, only the 2 positions of the length 10 is used, also there are 8 positions remaining, which is 28=256

If either begin or end with 1, then it would be 29 ways at beginning & 29 ways at ending respectively.

So required 10 bits will be generated in 28+29+29 = 256+512+512=3280

(J)

Since at least 1 men must be chosen, we consider all committees which include 4, 3, 2 , 1 and 5 men , with 1, 2, 3,4and 0 women, respectively.

Choosing 4 from 8 men and 1 from 7 women -> (8c4) * (7c1)

Choosing 3 from 8 men and 2 from 7 women-> (8c3) * (7c2)

Choosing 2 from 8 men and 3 from 7 women-> (8c2) * (7c3)

Choosing 1 from 8 men and 4 from 7 women-> (8c1) * (7c4)

Choosing 5 from 8 men and 0 from 7 women-> (8c5) * (7c0)

Total number of ways : (8c4) * (7c1)+ (8c3) * (7c2)+ (8c2) * (7c3)+ (8c1) * (7c4)+ (8c5) * (7c0)

= (70*7) + (56 * 21) +(28*35)+(8*35)+(56 * 1)=2982

(K) Considering all committees which include 1, 3, 4,0 men , with 3, 1, 0,4 women, respectively

10 C1 * 7c3 + 10 C3 * 7c1 + 10 c 4* 7c0 + 10 c 0 * 7c4

= 10* 35 + 120*7+210*1+1*35

= 1435

(L)

First putting all the numbers of pattern 7xxx (where x can be a number between 0 and 9). There is 1*9*9*9.

Then adding all the numbers of pattern y7xx (where y can be 1, 2, 3, 4, 5,6, 8, 9). There is 8*1*9*9

Then adding all the numbers of pattern yz7x (where z can be 0, 1, 2, 3, 4, 5,6, 8, 9). There is 8*9*1*9

Then adding all the numbers of pattern yzz7. There is 8*9*9*1

So in conclusion, (1*9*9*9) + (8*1*9*9) + (8*9*1*9) + (8*9*9*1) = 2673

(M)

The 5 problems are distinct and the 20 points as total score. Since we require each problem to be worth at least two point,. then need to assign 10 problems to 10 pointsfrom which there are

(10+10-1) C (10-1)

= 19 C 9

= 92378

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