Differentiation (Part-38) | S N De

In the previous article , we have discussed Short answer type questions from question no. 20-30 from Ex-7A (in the chapter Permutation &Combination). In this article, we will discuss rest of the solutions of Short Answer Type Questions from Ex-7A , Class XI as given in the Chhaya Publication Book or alternatively S N Dey mathematics solution class xi book of aforementioned chapter . So, let's start.

 

Short Answer Type Questions from Permutation Chapter | S N Dey Mathematics Class 11

$31.~~$ How many 6-digit even numbers can be formed from $~ 2,3,5,3,4,5~$ alone ?

Solution.

To form an even number , we have to place 2 or 4 at the unit place.

Now, after placing 2 at the unit place, rest of the 5 places must be filled with the given digits in which two digits (3 and 5) occur twice each and so the required arrangement is $~\frac{5!}{2! \times 2!}.$

Similarly, after placing 4 at the unit place, rest of the 5 places must be filled with the given digits in which two digits (3 and 5) occur twice each and so the required arrangement is $~\frac{5!}{2! \times 2!}.$

So, combining both possibility , total 6-digit even numbers that can be formed from $~ 2,3,5,3,4,5~$ alone are given by

$\frac{5!}{2! \times 2!}+\frac{5!}{2! \times 2!}=\frac{ 2 \times 5!}{2! \times 2!}=\frac{2 \times 120}{2 \times 2}=60.$

$32.~$ How many numbers lying between $~3000~$ and $~4000~$ can be formed with the digits $~0,1,2,3,4~$ , if repetitions of digits being allowed ?

Solution.

To find the numbers lying between $~3000~$ and $~4000~$ , the numbers should start with the digit $3$ and rest of the 3 places can be filled in by the given digits in 5 ways as repetitions of digits being allowed .

So, the total number of arrangement is $~1 \times 5 \times 5 \times 5=125.$ Out of these arrangement one number $3000$ will be generated which has to be excluded as we have to find number lying between $~3000~$ and $~4000~$.

So, the required arrangement is $~125-1=124.$

$33.~~$ In how many different ways can the four subjects be arranged in 5 periods of a class ?

Solution.

Since there are four subjects and 5 periods, so one subject will be taught twice. Hence, five subjects (with one subject repeating twice) can be arranged in 5 periods of a class in $~\frac{5!}{2!}~$ ways.

Again , four subjects can be selected in $4$ ways. Since for each subject repeating twice , the rest of the given subjects can be arranged in $~\frac{5!}{2!}~$ ways, the total number of ways can the four subjects be arranged in 5 periods of a class is given by :

$~\frac{5!}{2!} \times 4=\frac{120}{2} \times 4=240.$

$34.~~$ Find the rank of the word LATE when its letters are arranged as in a dictionary.

Solution.

$35.~~$ A code signal uses digit-letter-digit combinations but 0 and 1 are not used for either digits or letters. How many code signals are possible ?

Solution.

According to the problem, places need to be filled (or the number of ways in order to form a 3 digit number if Digits or letters can be repeated) = D L D.

Now, we notice that only 8 digits can be used to make a signal as 2 digits (0 and 1) are not allowed.

Letters that can be used to make a signal = 26 but 0 and 1 resemble with O and I. So, if we omit those two letters , rest of 26-2=24 letters can be used.

Also, letter or digits can be used repeatedly as we have to create a code (5 B 5 is different from 5 B 1)

Now, first place can be filled in = 8 ways whereas 2nd place can be filled in = 24 ways and 3rd place can be filled in = 8 ways.

Total no of ways to form a code = $8 \times 24 \times 8 = 1536.$