Differentiation (Part-38) | S N De

In the previous article, we have discussed VSA type questions 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.

Short Answer Type Questions for Permutation (Ex-7A) | S N Dey Mathematics | Class 11

1. If $ {}^{n+r}P_2=110 $ and $ {}^{n-r}P_2=20, $ find $ n $ and $ r.$

Solution

$ {}^{n+r}P_2=110 \\ \text{or, } \frac{(n+r)!}{(n+r-2)!}=110 \\ \text{or, } \frac{(n+r)(n+r-1)(n+r-2)!}{(n+r-2)!}=110 \rightarrow (1) $

$ {}^{n-r}P_2=20 \\ \text{or, } \frac{(n-r)!}{(n-r-2)!}=20 \\ \text{or, } \frac{(n-r)(n-r-1)(n-r-2)!}{(n-r-2)!}=20 \rightarrow (2) $

From $(1)$ we get:
$(n+r)(n+r-1) =110=11 \times 10 \rightarrow (3)$

From $(2)$ we get:
$ {(n-r)}(n-r-1) =20={5} \times 4 \rightarrow (4)$

From $(3)$ and $(4)$ we get:
$n+r=11 \rightarrow (5) ,~~ n-r=5 \rightarrow (6)$

From $(5)$ and $(6)$ we get:
$ (n+r)+(n-r)=11+5 \\ \text{or, } 2n=16 \\ \text{or, } n=\frac{16}{2}=8$

and so, by $(5)$, we get:
$r=11-n=11-8=3.$

2. In how many ways can the letters of the word LOGARITHM be arranged? How many of these arrangements begin with L? How many begin with L and do not end with M?

Solution

There are nine different letters in the word LOGARITHM and they can be arranged among themselves in $9!=362880$ ways.

Now to find the number of words beginning with L, we put the letter L in the first place. Then the remaining eight letters can be arranged among themselves in $8!=40320$ ways.

Finally, to find the number of words beginning with L and end with M, we put the letter L in the first place and M in the last place. Then the remaining seven letters can be arranged among themselves in $7!=5040$ ways.

Hence, the number of words that begin with L and do not end with M is obtained by the difference of the number of words beginning with L and the number of words beginning with L and end with M i.e.,

$8!-7!=40320-5040=35280.$

3. How many different arrangements of the letters of the word BENGAL can be made so that the two vowels do not come together?

Solution

There are six different letters in the word BENGAL and they can be arranged among themselves in $6!=720$ ways.

Now if two vowels E and A are taken as one letter, then the total number of letters is $(6-2+1)=5.$ These five letters can be arranged among themselves in $5!=120$ ways and for each of such arrangements, two vowels (E and A) can be arranged among themselves in $2!=2$ ways.

So, the number of words in which the two vowels always come together
$=120 \times 2=240.$

Clearly, the number of words in which the two vowels do not come together = (total number of words formed by letters in the word BENGAL) - (number of words in which the two vowels come together) = $720-240=480.$

So, the number of different arrangements of the letters of the word BENGAL can be made so that the two vowels do not come together is $480.$

4. In how many ways can the letters of the word STRANGE be arranged so that the vowels may appear in the odd places?

Solution

Clearly, there are seven letters in the word ‘STRANGE’ in which there are $2$ vowels. Also, there are $4$ odd places $(1,3,5,7)$ in the given word. So, we have to find the total number of ways for these four places which is ${}^4P_2=\frac{4!}{(4-2)!}=\frac{4 \times 3 \times 2!}{2!}=12.$

Corresponding to these $12$ ways the other $5$ letters may be placed in $5!=120$ ways and so total number of required arrangement is given by $12 \times 120=1440$ ways.

5. Find how many arrangements can be formed with the letters in the word JUXTAPOSED, the vowels always coming together.

Solution

Clearly, there are ten letters in the word ‘JUXTAPOSED’ in which there are $4$ vowels $(U, A, O, E)$. Now if four vowels are taken as one letter, then the total number of letters is $(10-4+1)=7.$ These seven letters can be arranged among themselves in $7!=5040$ ways and for each of such arrangements, four vowels can be arranged among themselves in $4!=24$ ways.

Hence, the number of arrangements which can be formed with the letters in the word JUXTAPOSED, the vowels always coming together is $5040 \times 24= 120960.$

6. Show that the number of ways in which $n$ books may be arranged on a shelf so that two particular books shall not be together is $(n-2)(n-1)!.$

Solution

If there is no restriction, $n$ books may be arranged on a shelf among themselves in $n!$ ways. If we take two particular books as one, then total number of books is $(n-2+1)=(n-1)$ which can be arranged among themselves in $(n-1)!$ ways. Again those two particular books can be arranged among themselves in $2!$ ways.

So, the number of ways in which a books may be arranged on a shelf so that two particular books shall be together is $2! \times (n-1)!.$

Therefore, the number of ways two particular books shall not be together = $n! - 2!(n-1)! = (n-1)![n - 2] = (n-2)(n-1)!.$

7. In how many ways $3$ boys and $5$ girls can be arranged in a row so that all the $3$ boys are together?

Solution

Incase, $3$ boys are always together, number of ways $3$ boys can be arranged is $3! = 6$.

The remaining $5$ girls must be arranged in $5$ places.
The number of ways $5$ girls and group of boys can be arranged in the row is $(5+1)!=6! = 720.$

Total number of ways boys and girls can be arranged in a row is $720 \times 6 = 4320.$

So, the boys and girls can be arranged in $4320$ ways.

8. In how many ways can $4$ boys and $3$ girls be arranged in a row so that no two girls come together?

Solution

Clearly there are $5$ places in between $4$ boys and $3$ girls can be arranged in $5$ places in ${}^5P_3$ ways whereas $4$ boys can be arranged among themselves in $4!$ ways.

Hence, number of ways can $4$ boys and $3$ girls be arranged in a row so that no two girls come together is

${}^5P_3 \times 4!=\frac{5!}{(5-3)!} \times 4!=\frac{5 \cdot 4 \cdot 3 \cdot 2!}{2!} \times 24=60 \times 24=1440$

9. How many arrangements can be made with all the letters of the word VENUS such that the order of the vowels remains unaltered?

Solution

If there is no restriction, $5$ letters in VENUS can be arranged among themselves in $5!$ ways.

So the number of arrangements in which the order of the vowels will remain unchanged (i.e. E will always precede U) is $\frac{5!}{2}=\frac{120}{2}=60.$

10. If none of the digits $2, 4, 5, 7, 8, 0$ be repeated, how many different numbers of $4$ digits can be formed with them?

Solution

If there is no restriction, then different $4$-digit numbers (taken out of $6$ digits) can be formed in ${}^6P_4$ ways.

Out of these $4$-digit numbers, there are numbers which starts with the digit $0$ and in that case rest of $3$ digits can be chosen in ${}^{6-1}P_{4-1}={}^5P_3$ ways.

So, in order to find $4$ digit numbers, we have to eliminate those numbers which start with the digit $0$ and so, the number of $4$ digits can be formed with them is

$={}^6P_4-{}^5P_3\\= \frac{6!}{(6-4)!}-\frac{5!}{(5-3)!}\\=\frac{6!}{2!}-\frac{5!}{2!}\\=6\cdot5 \cdot 4\cdot 3-5\cdot 4\cdot 3\\=360-60\\=300$