Problem 102 3 points Using the digits 1 2 3 and 5 how many 4

Problem 10.2 [3 points]. Using the digits 1, 2, 3 and 5, how many 4 digit numbers can be formed if (a) The first digit must be 1 and repetition of the digits s allowed? (b) The first digit must be 1 and repetition of the digits is not allowed? (c) The number must be divisible by 2 and repetion is allowed? (d) The number must be divisible by 2 and repetion is not allowed? (e) The number must be divisible by 2 or the first digit must be 1? Problem 10.3 [2 points]. Count the number of bit strings of length 10 such that: (a) There is no constraint. (b) The first bit must be 1. (c) The first bit must be 1 or the first two bits must be 01. (d) The sum of the bits must be even.

Solution

Ans 10.2(a):

first digit is occupied by 1 so only 3 places left.

we have 4 choice for each place left as repetition is allowed,

Hence there are total 1*4*4*4=64 numbers possible.

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Ans 10.2(b):

first digit is occupied by 1 so only 3 places left.

we have 3 choice for 2nd place as repetition is not allowed,

2 choice for 3rd place as repetition is not allowed,

1 choice for 4th place as repetition is not allowed,

Hence there are total 1*3*2*1=6 numbers possible.

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Ans 10.2(c):

Any number is divisible by 2 when unit digit is from 0,2,4,6,8.

that mens 4th digit has only one choice ( the number 2) from given numbers.

Since repeatition is allowed so each left place can be filled in 4 ways.

Hence there are total 4*4*4*1=64 numbers possible.

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Ans 10.2(d):

Any number is divisible by 2 when unit digit is from 0,2,4,6,8.

that mens 4th digit has only one choice ( the number 2) from given numbers.

Since repeatition is not allowed so each left place can be filled in 3,2,1 ways respectively; just like we did in part (b)

Hence there are total 3*2*1*1=6 numbers possible.

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Ans 10.2(e):

It doesn\'t say anything about repeatition so i will assume, repeatition is allowed.

we can do this part by adding possible numers for cases {number is divisible by 2} and {first digit is 1} then subtract case when {{number is divisible by 2 and first digit is 1 both} because this last case is common in first two cases.

Total numbers possible when number is divisible by 2= 64 {from part (c)}

Total numbers possible when first digit must be 1 = 64 {from part (a)}

Total numbers possible when number is divisible by 2 as well as first digit must be 1 =1*4*4*1=16

{here first and 4th digit has only one choice as explained before. 2nd and 3rd digits can be filled in 4 ways due to allowed repeatition}

Hence required possible numbers=64+64-16= 112 numbers