Differentiation (Part-38) | S N De

In the previous article , we have discussed Short Answer Type Questions (1-10) and solutions of Permutation . In this article, we will discuss the solutions of Short Answer Type Questions from Ex-7A in the chapter Permutation &Combination, Class XI as given in the Chhaya Publication Book or alternatively S N Dey mathematics solution class xi book of aforementioned chapter . So, without wasting time, let's start.

 

Permutation related problems & solutions | S N Dey mathematics solutions class xi

$11.~~$ How many numbers lying between $~3000~$ and $~4000~$ can be formed with the digits $~1, 2, 3, 4, 5~$ and $~6~$ ? (No repetition of digit is allowed.)

Solution.

By question, we will use the fact that all the numbers lying between $~3000~$ and $~4000~$ will have the digit $~3~$ at the thousands place. We will fix $~3~$ at the thousand's place. Then rest of $~5~$ digits can be arranged in $~3~$ places (hundred's, ten's and unit's place) in $~{}^5P_3~$ ways.

So, the numbers lying between 3000 and 4000 can be formed with the digits $~1, 2, 3, 4, 5~$ and $~6~$ is :

${}^5P_3=\frac{5!}{(5-3)!}=\frac{5!}{2!}=\frac{5\cdot 4\cdot 3 \cdot2!}{2!}=5\cdot 4\cdot 3=60~~\text{(ans.)}$

$12.~~$ If none of the digits 3, 5, 7, 8, 9 be repeated, how many different numbers greater than 7000 can be formed with them ?

Solution.

In this case, we have to find the number of integers greater than 7000 with the digits 3, 5,7,8 and 9. So, with these digits we can make maximum of five-digit numbers because repetition is not allowed.

Now, all the five-digit numbers are greater than 7000.

Therefore, the number of ways of forming 5-digit number = $5 \times 4\times 3\times2\times1 = 120$ and all the four-digit numbers greater than 7000 can be formed in the following manner.

Thousand place can be filled in 3 ways. Hundred place can be filled in 4 ways whereas tenth place can be filled in 3 ways. Units place can be filled in 2 ways.

Thus, we have total number of 4-digit number = $3 \times 4 \times 3\times 2 =72.$

$\therefore~$ Total number of integers $~120+72=192$.

$13.~~$ How many numbers of $4$-digit greater than $~6,000~$ can be formed with the digits $~3, 4, 5, 6, 8 ~~?$ (No digit is repeated in any number.) How many of these numbers so formed are odd?

Solution.

In order to form a number greater than $~6,000~$ with four digits out of the given five digits, $6$ or $8$ must be placed in the first place.

The number of arrangements starting with $6$ is $~{}^4P_3,~~$ as after starting with six, rest of three places must be filled up with $4$ digits $(3,4,5,8).$

Similarly, the number of arrangements starting with $8$ is $~{}^4P_3.$

Hence, numbers of $4$-digit greater than $~6,000~$ can be formed with the digits $~3, 4, 5, 6, 8 ~~$ are

${}^4P_3+{}^4P_3\\~=2 \times \frac{4!}{(4-3)!}\\~=2 \times \frac{4.3.2.1!}{1!}\\~=48$

To find the odd numbers unit place must be placed with $3$ or $5$ whereas thousands place must be filled with $6$ or $8$.

So, total number of arrangement for the middle two places out of 3 numbers with 6 and 3 at the beginning and at the end is $~~~{}^3P_2.$

Similarly, total number of arrangement for the middle two places out of 3 numbers with 6 and 5 at the beginning and at the end is $~~~{}^3P_2.$

Again, total number of arrangement for the middle two places out of 3 numbers with 8 and 3 at the beginning and at the end is $~ {}^3P_2.$

Finally, total number of arrangement for the middle two places out of 3 numbers $(3,4,6)$ with $8$ and $5$ at the beginning and at the end is $~~~{}^3P_2.$

Hence, total odd numbers formed

${}^3P_2+{}^3P_2+{}^3P_2+{}^3P_2\\~=4 \times {}^3P_2 \\~=4 \times \frac{3!}{(3-2)!}\\~= 4 \times 3!~~[\because~ 1!=1]\\~=4!\\~=24.$

$14.$ How many numbers of $5$ digits can be formed with the digits $0, 2, 5, 6, 7$ without taking any of these digits more than once ?

Solution.

If there is no restriction, digits $0, 2, 5, 6, 7$ can be arranged among themselves in $~{}^5P_5=5!~$ ways, out of which there are numbers starting with $~0~$ and with $~0~$ at the beginning, rest of the digits can be arranged among themselves in $~{}^4P_4=4!~$ ways. Theses are $~4~$ digit numbers and so they must be deducted.

So, total numbers of $5$ digits can be formed with the digits $0, 2, 5, 6, 7$ without taking any of these digits more than once , are

$~5!-4!=120-24=96.$

15. How many odd numbers of $5$ significant digits can be formed with the digits $~3, 6, 7, 2, 0~$ when no digit is repeated? [Council Sample Question '13]

Solution.

With the given digits, the number could be odd only if it ends with either $3$ or $7$. Let us first consider $7$ as the fixed last digit and try to find the possible numbers ending with $7$.

Since it is a $5$ digit number and so $0$ can never be placed at the first place and that is the reason why we are left with three ways $(3,6,2)$ to fill the first digit.

Now for the place of second digit we can include $0$ too and again we can fill the second place by three ways (because the digits aren't allowed to repeat).

Finally, we are left with two ways to fill the third place digit and just one way to fill the fourth place digit.

Hence, the total no. of ways are: $~3 \times 3 \times 2\times 1=18.$

Similarly, when we take the last digit to be 3 ,we get the same result.

Therefore, total possible numbers are $18+18=36$.

$16.~$ In how many ways can 5 commerce and 4 science students can be arranged in a row so that the commerce and the science students are placed alternatively.

Solution.

There are $4+1=5$ vacant places among $4$ science students.

According to the problem, $5$ commerce students are to be placed in the those $5$ vacant places so that the commerce and the science students can be placed alternatively.

Now, 5 commerce students can be arranged in $~{}^5P_5=5!~$ ways whereas with each of such $5!$ arrangements, $4$ science students can arrange among themselves in $~{}^4P_4=4!~$ ways.

Hence, total number of such arrangement is $~5! \times 4!=120 \times 24=2880~$ so that the commerce and the science students can be placed alternatively.

$17.~$ In how many ways can 5 first year students and 3 second year students be seated in a row so that no two second year students may sit together ?

Solution.

There are $5+1=6$ vacant places among $5$ first year students. According to the problem, $3$ second year students are to be placed in the those $6$ vacant places so that the first and the second year students can be placed alternatively.

Now, $3$ second year students can be arranged in $~{}^6P_3=\frac{6!}{3!}~$ ways whereas with each of such arrangements, $5$ first year students can arrange among themselves in $~{}^5P_5=5!~$ ways.

Hence, total number of such arrangement is

$~\frac{6!}{3!} \times 5!=\frac{6 \times 5!}{6} \times 120=120 \times 120=14400~$

and with such arrangements the first and the second year students may be seated in a row so that no two second year students may sit together.

$18.~$ In how many of the permutations of $~12~$ things taken $~3~$ at a time will one particular thing $(a)~$ always occur $~(b)~$ never occur.

Solution.

We know that total permutation of $n$ different things taken $r$ at a time when a particular item is always included in the arrangement is given by :

$r \times {}^{n-1}P_{r-1}\\~~~=3 \times {}^{12-1}P_{3-1}\\~~~=3 \times {}^{11}P_{2}\\~~~=3 \times \frac{11!}{(11-2)!}\\~~~=3 \times \frac{11 \cdot 10 \cdot 9!}{9!}\\~~~=3 \times 11 \times 10\\~~~=330.$

If out of $12$ different thing one thing will never occur ,then $3$ things will occur out of $~(12-1)=11~$ things.

$\therefore~$ the required arrangement in this case is :

${}^{11}P_3=\frac{11!}{(11-3)!}=\frac{11!}{8!}=\frac{11 \cdot 10 \cdot 9 \cdot 8!}{8!}=990.$

$19.~$ In many of the permutations of $12$ things taken $6$ at a time will $3$ particular things $~(a)~$ always occur $~(b)~$ never occur ?

Solution.

According to the problem, if we keep those $3$ particular things aside , then rest of $~(6-3)=3~$ things are to be picked from $~(12-3)=9~$ things.

Now, $~3~$ things can be picked from $~6~$ things in $~{}^6P_3~$ ways and for each of such arrangement , rest of $~3~$ things can be selected from $~9~$ things in $~{}^9P_3~$ ways.

So, required number of arrangement such that $3$ particular things always occur is :

$ {}^6P_3 \times {}^9P_3=\frac{6!}{(6-3)!} \times \frac{9!}{(9-3)!}=\frac{6! \times 9!}{3! \times 6!}=\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3!}{3!}=60480.$