Advertisements
Advertisements
Question
A test consists of 10 multiple choice questions. In how many ways can the test be answered if the first four questions have three choices and the remaining have five choices?
Advertisements
Solution
The first four questions have 3 choices.
So they can be answered in 34 ways.
The remaining 6 questions have 5 choices.
So they can be answered in 56 ways.
So all 10 questions can be answered in 34 × 56 ways.
APPEARS IN
RELATED QUESTIONS
Compute `(8!)/(6! xx 2!)`
Evaluate `(n!)/((n-r)!)` when n = 6, r = 2
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
In how many ways can the letters of the word PERMUTATIONS be arranged if the vowels are all together.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Evaluate each of the following:
6P6
In how many ways can 4 letters be posted in 5 letter boxes?
In how many ways 4 women draw water from 4 taps, if no tap remains unused?
Write the number of all possible words that can be formed using the letters of the word 'MATHEMATICS'.
Write the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together ?
The number of ways in which the letters of the word ARTICLE can be arranged so that even places are always occupied by consonants is
Suppose 8 people enter an event in a swimming meet. In how many ways could the gold, silver and bronze prizes be awarded?
8 women and 6 men are standing in a line. In how many arrangements will all 6 men be standing next to one another?
There are 10 persons named P1, P2, P3, ... P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.
Let b1, b2, b3, b4 be a 4-element permutation with bi ∈ {1, 2, 3, .......,100} for 1 ≤ i ≤ 4 and bi ≠ bj for i ≠ j, such that either b1, b2, b3 are consecutive integers or b2, b3, b4 are consecutive integers. Then the number of such permutations b1, b2, b3, b4 is equal to ______.
If 1P1 + 2. 2p2 + 3. 3p3 + ....... 15. 15P15 = qPr – s, 0 ≤ s ≤ 1, then q+sCr–s is equal to ______.
The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is ______.
If m+nP2 = 90 and m–nP2 = 30, then (m, n) is given by ______.
