Differentiation (Part-38) | S N De

In the previous article , we have discussed Short Answer Type Questions (11-19) 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 | Short Answer Type Questions | S N Dey mathematics Class 11

$20.~$ In how many different ways can the letters of the word FOOTBALL be arranged so that two O's do not come together ?

Solution.

In the given 8 lettered word , O and L appear twice and so the total number of arrangements of the letters $~\frac{8!}{2! \times 2!}.$

If two O's are together or we consider them as one letter , then total number of letters in FOOTBALL is : $ (8-2+1)=7 ~$ and so in that case total number of arrangement of letters is $~\frac{7!}{2!}. $

Finally, number of ways can the letters of the word FOOTBALL be arranged so that two O's do not come together is

$~~\frac{8!}{2! \times 2!}-\frac{7!}{2!}\\~=\frac{8.7!}{2 \times 2}-\frac{7!}{2}\\~=\frac{7!}{2} \times (\frac{8}{2}-1)\\~=7560.$

$21.~$ How many different arrangements are possible with the letters of the word ALGEBRA ? In how many of these , two A's will not be together ?

Solution.

In the given 7 lettered word , A appears twice and so the total number of arrangements of the letters $~\frac{7!}{2!}=2520.$

If two A's are together or we consider them as one letter , then total number of letters in ALGEBRA is : $ (7-2+1)=6 ~$ and so in that case total number of arrangement of letters is $~6!=720. $

So, the number of ways in which two A's will not be together is

$~ 2520-720=1800. $

$22.~~$ How many different words can be formed taking all the letters in the word PEOPLE in which two P's will not come together ?

Solution.

In the given 6 lettered word , letters P, E appear twice and so the total number of arrangements of the letters $~\frac{6!}{2! \times 2!}$

If two P's are together or we consider them as one letter , then total number of letters in PEOPLE is : $ (6-2+1)=5 ~$ and so in that case total number of arrangement of letters is $~\frac{5!}{2!}. $

So, finally number of arrangements in which two P's will not come together is

$~~\frac{6!}{2! \times 2!}-\frac{5!}{2!}\\~=\frac{6.5!}{2 \times 2}-\frac{5!}{2}\\~=\frac{5!}{2} \times (\frac{6}{2}-1)\\~=120$

$23.~~$ How many arrangements can be made out of the letters of the word COMMITTEE at a time, such that the four vowels do not come together ?

Solution.

In the given 9 lettered word , letters M,T,E appear twice and so the total number of arrangements of the letters $~\frac{9!}{2! \times 2! \times 2!}. $

If four vowels (O,I,E,E) come together or we consider them as one letter , then total number of letters in COMMITTEE is : $ (9-4+1)=6 ~$ and so in that case total number of arrangement of these 6 letters is $~\frac{6!}{2! \times 2!},~ $ as in this 6 lettered word M,T appear twice. Again four vowels can be arranged among themselves in $~\frac{4!}{2!}~$ ways as in four vowels, E appear twice.

So, the number of arrangements in which four vowels come together is

$\frac{6!}{2! \times 2!} \times \frac{4!}{2!}=2160.$

Finally, total number of arrangements in which the four vowels do not come together , is given by

$~\frac{9!}{2! \times 2! \times 2!}-2160=45360-2160=43200. $

$24.~~$ Find the number of ways in which the letters of ORION can be arranged so that two constants do not come together.

Solution.

In the given 5 lettered word , the letter O appears twice and so the total number of arrangements of the letters $~\frac{5!}{2!}. $

If two constants (R, N) come together or we consider them as one letter , then total number of letters in ORION is : $ (5-2+1)=4 ~$ and so in that case total number of arrangement of these 4 letters is $~\frac{4!}{2! },~ $ as in this 4 lettered word O appears twice. Again two constants can be arranged among themselves in $~2!~$ ways .

So, the number of arrangement in which two constants come together, is given by

$\frac{4!}{2! } \times 2!=4!=24.$

Hence, the number of arrangement in which two constants do not come together, is given by

$\frac{5!}{2!}-24=\frac{120}{2}-24=60-24=36.$

$25.~$ Show that the letters of the word INSURANCE can be arranged in twice as many ways as the letters of the word ECONOMICS.

Solution.

In the given 9 lettered word INSURANCE , the letter N appears twice and so the total number of arrangements of the letters $~\frac{9!}{2!} \rightarrow(1)$

Again, in the given 9 lettered word ECONOMICS , the letters C,O appear twice and so the total number of arrangements of the letters $~\frac{9!}{2! \times 2!}=\frac12 \times \frac{9!}{2!} \rightarrow(2)$

From $(1),(2)~$ we can conclude that the letters of the word INSURANCE can be arranged in twice as many ways as the letters of the word ECONOMICS.

$26.~~$ How many different arrangements can be made out of the letters of the expression $~x^3y^2z^4~$ when written at full length ?

Solution.

In the given 9 lettered expression $~x^3y^2z^4~$ , the letter $x$ appears thrice, the letter $y$ appears twice and the letter $z$ appears $4$ times . So, the total number of arrangements of the letters $~\frac{9!}{3! \times 2! \times 4!}=\frac{ 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4!}{ 6\times 2 \times 4!}=1260.$

$27.~~$ The first name of a person consists of 9 letters in which one letter occurs more than once and the other letters are different. If the number of permutations of the letters of his name taken all at a time be 15120, find the number of times the like letter occurs.

Solution.

Since the first name of a person consists of 9 letters in which one letter occurs $x$ times (say).

Then according to the problem,

$ \frac{9!}{x!}=15120 \\ \text{or,}~~ x!=\frac{9!}{15120} \\ \text{or,}~~ x! =\frac{9 \cdot 8 \cdot7!}{15120} \\ \text{or,}~~ x!=\frac{7!}{210} \\ \text{or,}~~ x!=24=4! \\ \text{or,}~~x=4. $

$28.~~$ How many numbers of six digits can be formed out of the digits of the number 567724? How many of these numbers so formed are even?

Solution.

$29.~~$ How many numbers of 8 digits can be formed with the digits $~4, 2, 2, 2, 3, 3, 5, 5~$ so that odd digits are always in odd places?

Solution.

In a 8-digit number there are 4 odd places and 3 even places. Now, according to the problem, there are 4 even numbers $(4,2,2,2)$ and 4 odd numbers $(3, 3, 5, 5).$

So, there can be $\frac{4!}{2! \times 2!}=6~$ arrangements by keeping odd digits in odd places, as the digits $3,~5~$ occur twice.

Again, there can be $\frac{4!}{3!}=4~$ arrangements by keeping even digits in even places, as the digit $~2~$ occurs twice.

Hence , there are total of $~6 \times 4=24~$ numbers of 8 digits which can be formed with the digits $~4, 2, 2, 2, 3, 3, 5, 5~$ so that odd digits are always in odd places.

$30.~~$ How many numbers greater than a lac be formed with the digits $ 0, 2, 5, 2, 4, 5 ?$

Solution.

According to the problem, there are $6$ digits in which the digits $~2~$ and $~5~$ appear twice and so the total number of arrangements is given by $~\frac{6!}{2! \times 2!}=180.$

Again, placing $~0~$ at start, rest of the digits can be arranged in $~\frac{5!}{2! \times 2!}=30~$ ways and the numbers formed in this arrangement is less than one lac.

So, total numbers greater than a lac be formed with the digits $ 0, 2, 5, 2, 4, 5 ~$ are given by

$180-30=150.$