Test: Single Correct MCQs: Permutations and Combinations | JEE Advanced
15 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Permutations and Combinations | JEE Advanced
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nCr–1 = 36, nCr = 84 and nCr + 1 = 126, then r is : (1979)
nCr -1 = 36, nCr = 84,nCr+1= 126
KEY CONCEPT : We know that
⇒ 
....(1)
⇒ 
⇒ 2n – 5r – 3 = 0 ....(2)
Solving (1) and (2), we get n = 9 and r = 3.
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are (1982 - 2 Marks)
Total number of words that can be formed using 5 letters out of 10 given different letters
= 10 × 10 × 10 × 10 × 10 (as letters can repeat)
= 1, 00, 000
Number of words that can be formed using 5 different letters out of 10 different letters
= 10P5 (none can repeat)
∴ Number of words in which at least one letter is repeated
= total words–words with none of the letters repeated
= 1,00,000 – 30,240
= 69760
The value of the expression 
is equal to (1982 - 2 Marks)
is equal to (1982 - 2 Marks)
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is (1982 - 2 Marks)
Two women can choose two chairs out of 1, 2, 3, 4, in 4C2 ways and can arrange themselves in 2! ways.
Three men can choose 3 chairs out of 6 remaining chairs in 6C3 ways and can arrange themselves in 3! ways
∴ Total number of possible arrangements are
4C2× 2! × 6C3 × 3!=4P2× 6P3
A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is (1989 - 2 Marks)
KEY CONCEPT : We know that a number is divisible by 3 if the sum of its digits is divisibly by 3.
Now out of 0, 1, 2, 3, 4, 5 if we take 1, 2, 3, 4, 5 or 0, 1, 2, 4, 5 then the 5digit numbers will be divisible by 3.
Case I : Number of 5 digit numbers formed using the
digits 1, 2, 3, 4, 5 = 5! = 120
Case II : Taking 0, 1, 2, 4, 5 if we make 5 digit number then
I place can be filled in = 4 ways (0 can not come at I place)
II place can be filled in = 4 ways
III place can be filled in = 3 ways
IV place can be filled in = 2 ways
V place can be filled in = 1 ways
∴ Total numbers are = 4 × 4! = 96
Thus total numbers divisible by 3 are
= 120 + 96 = 216
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions ? (2000S)
X – X – X – X – X. The four digits 3, 3, 5, 5 can be arranged at (–) places in 
= 6 ways.
The five digits 2, 2, 8, 8, 8 can be arranged at
(X) places in 
= 10ways.
Total no. of arrangements = 6 × 10 = 60 ways.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn + 1 - Tn = 21, then n equals (2001S)
As per question,
⇒ n(n - 1) (n + 1 -n + 2)= 126
⇒ n (n – 1) = 42 ⇒ n (n – 1) = 7 × 6 ⇒ n = 7.
The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently is (2002S)
Total number of ways of arranging the letters of the word BANANA is 
= 60 Number of words in which 2 N’s come together is 
= 20
Hence the required number = 60 – 20 = 40
A rectangle with sides of length (2m – 1) an d (2n – 1) units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side length s is (2005S)
If we see the blocks in terms of lines then there are 2m vertical lines and 2n horizontal lines. To form the required rectangle we must select two horizontal lines, one even numbered (out of 2, 4, .....2n) and one odd numbered (out of 1, 3....2n–1) and similarly two vertical lines. The number of rectangles is
mC1 . mC1 . nC1 . nC1 = m2n2
If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)
∵ r, s, t are prime numbers,
∴ Section of (p, q) can be done as follows
∴ r can be selected 1 + 1 + 3 = 5 ways
Similarly s and t can be selected in 9 and 5 ways respectivley.
∴ Total ways = 5 × 9 × 5 = 225
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is (2007 -3 marks)
The letter of word COCHIN in alphabetic order are C, C, H, I, N, O.
Fixing first letter C and keeping C at second place, rest 4 can be arranged in 4! ways.
Similarly the words starting with CH, CI, CN are 4! in each case.
Then fixing first two letters as CO next four places when filled in alphabetic order give the word COCHIN.
∴ Numbers of words coming before COCHIN are
4 × 4! = 4 × 24 = 96
The number of seven digit i ntegers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is (2009)
We h ave to for m 7 digit n umber s, using th e digits 1, 2 and 3 only, such that the sum of the digits in a number = 10.
This can be done by taking 2, 2, 2, 1, 1, 1, 1, or by taking 2, 3, 1, 1, 1, 1, 1.
∴ Number of ways
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is (2012)
∵ Each person gets at least one ball.
∴ 3 Per son s can have 5 balls in th e followin g systems
or
The number of ways to distribute the balls in first system =
5C1 x 4C1 x 3C3
Also 3, persons having 1, 1 and 3 balls can be arranged in 
ways.
∴ No. of ways to distribute 1, 1, 3 balls to the three persons
5C1 x 4C1 x 3C3 x
Similarly the total no. of ways to distribute 1, 2, 2 balls to the three persons = 5C1 x 4C2 x 2C2 x
∴ The required number of ways = 60 + 90 = 150
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is (JEE Adv. 2014)
∵ Car d numbered 1 is always placed in envelope numbered 2, we can consider two cases.
Case I: Card numbered 2 is placed in envelope numbered 1.
Then it is dearrangement of 4 objects, which can be done in
= 9 ways
Case II: Card numbered 2 is not placed in envelope numbered 1.
Then it is dearrangement of 5 objects, which can be done in
= 44 ways
∴ Total ways = 44 + 9 = 53
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 memoers) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is (JEE Adv. 2016)
Either one boy will be selected or no boy will be selected. Also out of four members one captain is to be selected.
∴ Required number of ways = (4C1 × 6C3+ 6C4) × 4C1 = (80 + 15) × 4 = 380
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