JEE Questions for Maths Permutations And Combinations Quiz 8 - MCQExams.com

Fix up 1 man and the remaining 6 men can be seated in 6! ways. Now  no. two women are to sit together and as such the 7 women are to be arranged in seven empty seats between two consecutive men and number of arrangement will be 7! . Hence by fundamental theorem the total number of ways = 7! × 6! .

There are 20 + 1 = 21 persons in all. The two particular persons and the host be taken as one unit so that these remain 21 – 3 + 1 = 19 persons to be arranged in 18! ways. But the two person on either side of the host can themselves be arranged in 2!  ways.Hence there are 2!18! ways or 2.18! ways.

Total number of ways to distribute one  Rs. 100 note and five other notes  = 3⁶.

Given set of numbers is {1,2,…….11} in which 5 are even six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.

Words starting with A, C, H, I, N are each equals to 5!

∴ Total words  = 5 × 5! = 600

The first word starting with S is SACHIN.

 ∴ SACHIN appears in dictionary at serial number 601.

Total number of arrangements of n books = n!

If two specified books always together then number of ways = (n – 1)! × 2

Hence required number of ways = n! – (n – 1)! × 2

= n(n – 1)! – (n – 1)! × 2 = (n – 1)! (n – 2).

The last place can be filled by 1,3,5,7 as the number is to be odd. Therefore, the digit at the extreme right can be filled by 4 ways. The digit at the extreme left can be filled by 5 ways as zero has to be excluded and the remaining II and III positions can be filled by 6 ways each. Hence total number of ways are 5 × 6 × 6 × 4 = 720.

Numbers greater than 1000 and less than or equal to 4000 will be of 4 digits and will have either 1 (except 1000) or 2 or 3 in the  first place with 0 in each of remaining places.

After fixing 1^st place , the second place can be filled by any of the 5 numbers. Similarly third place can be filled up in 5 ways and 4^th place can be filled up in 5 ways. Thus there will be 5 × 5 × 5 = 125 ways in which 1 will be in first place but this include 1000 also hence there will be 124 numbers having 1 in the first place. Similarly 125 for each 2 or 3. One number will be in which 4 in the first place and i.e., 4000. Hence the required numbers are 124 + 125 + 125 + 1 = 375ways.

Since first digit cannot be zero. Total no. of 5 digit number = 9 × 10 × 10 × 10 × 10 = 90,000 and no. of 5 digit number which have none of the digits repeated

= 9 × 9 × 8 × 7 × 6 = 27,216. Numbers of  5 digit no. with repeated digit = 90,000 – 27,216 = 62,784.

Out of 7 places , 4 places are odd and 3 even . Therefore 3 vowels can be arranged in 3 even places in ³P₃ ways and remaining 4 consonants can be arranged in 4 odd places in ⁴P₄ ways.Hence required no. of ways = ³P₃ × ⁴P₄ = 144.

At first we have to accommodate those  5 animals  in cages which can not enter in 4 small cages, therefore, number of ways are ⁶P₅. Now after accommodating 5 animals we left with 5 cages and 5 animals , therefore numbers of ways are 5!.Hence required number of ways  = ⁶P₅ × 5! = 86400.

If 1 is at 1000^th, 100^th, 10^th  and unit place then the number of four digit, no. will be 7 × 6 × 5, 6 × 6 × 5, 6 × 6 × 5,  6 × 6 × 5 respectively. So the total four digit number

= 7 × 6 × 5 + 6 × 6 × 5 + 6 × 6 × 5 + 6 × 6 × 5 = 750.

3 must be at thousand place and since the number should be divisible by 5,so 5 must be at unit place. Now we have to fill two place (ten and hundred) i.e., ⁴P₂ = 12.

As books of  same subject are together, we can think of there bundles. Now these three bundles can be arranged in 3! Ways. Also, each of them can be arranged internally to make a different arrangement. Hence , total ways = 3! × 5! × 4! × 2! = 5! 4! 3! 2!.

5 boys can be stand in a row 5! Ways

Now , two girls can’t stand  in a row together in ⁶P₅ ways.

Total no. of required arrangement = 5! × ⁶P₅ = 5! × 6!

Statement 1 :
B₁ + B₂ + B₃ + B₄ = 10
= Coefficient of x¹⁰ in (x¹ + x² + ....+ x⁷)⁴
= Coefficient of x⁶ in (1 - x⁷)⁴ (1 - x)^-4
= ^4+6-1C₃ = ⁹C₃.
Statement 2 : Obviously ⁹C₃.

x₁ + x₂ + x₃ + x₄ + x₅ = 6
^5+6-1C_5-1 = ¹⁰C₄.

Since at any place, any of the digits, 2,5 and 7 can be used, total number of such positive n-digit numbers are 3ⁿ.Since we have to form 900 distinct numbers, hence 3ⁿ ≥ 900 ⇒ n = 7.

The required number of ways  = ⁸C₃ – ⁵C₃ – ³C₃.

[Since total points are 8, but 5 are collinear and other three are also collinear]